Consider the vaporization of liquid water to steam at a pressure of 1 atm. In what temperature range is it a spontaneous process?

Answers

Answer 1

The vaporization of liquid water to steam is a spontaneous process when the temperature is above its boiling point at a pressure of 1 atm.

At 1 atm, the boiling point of water is 100 degrees Celsius or 212 degrees Fahrenheit. When the temperature of the water reaches or exceeds this value, the average kinetic energy of the water molecules increases, allowing more molecules to overcome the intermolecular forces and escape into the gas phase. At temperatures below the boiling point, water molecules have lower average kinetic energy, and the intermolecular forces between the water molecules are stronger. In this case, the rate of evaporation is slower, and the process is non-spontaneous. However, it is important to note that even at temperatures below the boiling point, water molecules with higher kinetic energy can still evaporate from the surface. Above the boiling point, the kinetic energy of the water molecules is sufficient to overcome the intermolecular forces completely, and the evaporation process becomes more rapid and spontaneous. The water molecules have enough energy to transition into the gas phase without requiring additional external energy input. It is crucial to maintain a temperature within the appropriate range to ensure a spontaneous vaporization process. If the temperature falls below the boiling point, the process may slow down or stop altogether. Conversely, if the temperature rises significantly above the boiling point, it may result in superheating, where the water remains in the liquid state despite being above the boiling point. This can be unstable and potentially lead to a sudden, explosive boiling known as a "bump." Therefore, maintaining a temperature within the range of the boiling point of water at 1 atm pressure ensures a spontaneous and controlled vaporization process.

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Related Questions

A typical airbag in a car is 140 liters. How many grams of sodium azide needs to be loaded into an airbag to fully inflate it at standard temperature and pressure? (see the chemical reaction from the introduction). For credit, please show all work. 405.85 grams

Answers

Therefore, 237.931 grams of sodium azide would be needed to fully inflate a typical car airbag at standard temperature and pressure

To determine the amount of sodium azide needed to fully inflate a typical car airbag, we need to consider the stoichiometry of the chemical reaction involved. The reaction between sodium azide (NaN₃) and a metal oxide, typically potassium nitrate (KNO₃), produces nitrogen gas (N₂) and sodium oxide (Na₂O):

2 NaN₃ + 2 KNO₃→ 3 N₂ + 2 Na₂O + K₂O

According to the reaction equation, 2 moles of NaN₃ produce 3 moles of N₂. We need to find the number of moles of sodium azide required to fill a 140-liter airbag.

First, we need to convert the volume of the airbag from liters to moles of nitrogen gas. To do this, we'll use the ideal gas law:

PV = nRT

Where:

P = Pressure (standard pressure = 1 atm)

V = Volume (in liters)

n = Number of moles

R = Ideal gas constant (0.0821 L·atm/mol·K)

T = Temperature (standard temperature = 273.15 K)

Using the values:

P = 1 atm

V = 140 liters

R = 0.0821 L·atm/mol·K

T = 273.15 K

We can rearrange the ideal gas law to solve for the number of moles (n):

n = PV / RT

n = (1 atm) * (140 L) / (0.0821 L·atm/mol·K * 273.15 K)

n = 5.4956 moles of N2

Now, since the stoichiometry of the reaction tells us that 2 moles of NaN₃ produce 3 moles of N₂, we can set up a proportion:

2 moles NaN₃ / 3 moles N₃ = x moles NaN₃ / 5.4956 moles N₂

Simplifying the proportion, we find:

x = (2 moles NaN₃ * 5.4956 moles N₂) / 3 moles N₂

x = 3.6637 moles NaN₃

Finally, to convert moles of sodium azide to grams, we need to multiply by the molar mass of NaN₃, which is approximately 65 grams/mol:

Mass of NaN₃ = 3.6637 moles * 65 g/mol

Mass of NaN₃ = 237.931 grams

Therefore, 237.931 grams of sodium azide would be needed to fully inflate a typical car airbag at standard temperature and pressure.

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the eiffel tower is a steel structure whose height increases by 19.8 cm when the temperature changes from -9 to +41 oc. what is the approximate height (in meters) at the lower temperature?

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The approximate height of the Eiffel Tower at the lower temperature is 0.33 meters.

To determine the approximate height of the Eiffel Tower at the lower temperature, we can use the formula for linear thermal expansion;

ΔL = αLΔT

where; ΔL is the change in length

α is the coefficient of linear expansion

L is the original length

ΔT is the change in temperature

Given;

ΔL = 19.8 cm = 0.198 m (converted to meters)

ΔT = (+41 °C) - (-9 °C) = 50 °C

α will be the coefficient of linear expansion for steel.

The coefficient of linear expansion for steel will be approximately 12 x 10⁻⁶ °C⁻¹.

Using the formula, we can solve for the original length (L);

0.198 m = (12 x 10⁻⁶ °C⁻¹ × L × 50 °C

Simplifying the equation;

L = 0.198 m / (12 x 10⁻⁶ °C⁻¹ × 50 °C)

L ≈ 0.33 meter

Therefore, the approximate height of the Eiffel Tower will be 0.33 meter.

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use the chemical equation to answer the question. 2h2(g) o2(g) → 2h2o(l) which statement describes the breaking and forming of bonds in the reaction?

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In the chemical equation 2H2(g) + O2(g) → 2H2O(l), the breaking of bonds occurs in the reactants (H2 and O2), and the formation of bonds takes place in the product (H2O). The reaction involves the breaking of the H-H bond in hydrogen molecules (H2) and the O=O bond in oxygen molecules (O2). These bonds are relatively weak and require energy input to break. On the other hand, the formation of the H-O bonds in water molecules (H2O) releases energy as new bonds are formed. This process involves the sharing of electrons between hydrogen and oxygen atoms to form covalent bonds.

In the given chemical equation, 2H2(g) + O2(g) → 2H2O(l), the breaking and forming of bonds occur during the reaction. The reactants consist of hydrogen molecules (H2) and oxygen molecules (O2), and the product is water (H2O).

In the reactants, each hydrogen molecule contains a covalent bond between the two hydrogen atoms, known as the H-H bond. Similarly, each oxygen molecule has a covalent bond between the two oxygen atoms, called the O=O bond. These bonds are relatively weak and require energy input to break.

During the reaction, the H-H bonds and O=O bonds in the reactants are broken. Energy is supplied to break these bonds, which allows the hydrogen and oxygen atoms to become more reactive. The breaking of bonds is an endothermic process because it requires energy input.

As the reaction progresses, new bonds are formed in the product, which is water (H2O). Water molecules contain covalent bonds between hydrogen and oxygen atoms, known as the H-O bonds. The formation of these bonds releases energy, making the process exothermic.

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In photorespiration, release of CO
2
occurs in
A
chloroplast
B
mitochondria
C
peroxisomes
D
glyoxysome

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In photorespiration, the release of CO₂ occurs in peroxisomes. Photorespiration is the process where O₂ is taken in, and CO₂ is released during photosynthesis. (option.c)

However, photorespiration is a wasteful process as it inhibits photosynthesis rather than helping it. It takes place in the chloroplast and peroxisomes. The oxygen concentration inside the leaf is high, which leads to an oxygenation reaction taking place instead of a carboxylation reaction.

Photorespiration occurs in plants that have adapted to hot and dry environments, or in plants that are still developing their photosynthetic capacity. It is also known as the C₂ cycle, which involves the breakdown of a compound called glycolate.

Glycolate is produced when RuBisCO (an enzyme) reacts with O instead of CO₂. When glycolate is broken down, CO₂ is produced in the₂ peroxisome, and then enters the Calvin cycle for photosynthesis.The glyoxysome is a type of peroxisome found in the cells of plants and fungi.

It is involved in the conversion of stored lipids into carbohydrates, but it is not involved in photorespiration. Hence, CO₂ is released in peroxisomes during photorespiration.

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the theoretical van't hoff factor means that total particle concentration

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The theoretical Van't Hoff factor means that the total particle concentration. A solution is a homogenous mixture of two or more substances.

It is composed of a solute and a solvent. A solute is the substance that dissolves in a solvent. Solvent is a substance in which other substances dissolve. The number of ions formed when a solute dissolves in a solvent is calculated using the Van't Hoff factor.

When a solute dissolves in a solvent, the number of particles in the solution increases. The degree of dissociation of an ionic compound in water can be determined using the Van't Hoff factor.

The theoretical Van't Hoff factor is the number of ions that would be present if all of the solute dissolved. It is determined by dividing the measured molality by the expected molality of the solute.

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Write a C++ program to keep asking for a number until you enter a negative number. At the end, print the sum of all entered numbers. You need to use a while loop for this Lab.

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For a C++ program to keep asking for a number until you enter a negative number, Here's a C++ program that meets the requirements:

How to write the program

#include <iostream>

int main() {

   int number;

   int sum = 0;

   

   std::cout << "Enter numbers (enter a negative number to exit):\n";

   

   while (true) {

       std::cout << "Enter a number: ";

       std::cin >> number;

       

       if (number < 0) {

           break; // Exit the loop when a negative number is entered

       }

       

       sum += number;

   }

   

   std::cout << "Sum of all entered numbers: " << sum << std::endl;

   

   return 0;

}

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Design a procedure that would test the law of conservation of mass for a burning log

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Procedure for testing the law of conservation of mass for a burning log :When a log burns, the law of conservation of mass must be tested.

The procedure is as follows:

Step 1: Obtain a log of known mass and record it as M1.

Step 2: Burn the log completely and collect the ash left behind.

Step 3: Record the mass of the ash as M2.

Step 4: Determine the mass of the combustion product released into the air by subtracting the mass of the ash from the original mass of the log. This would be M3 = M1 – M2.

Step 5: Use a balance to weigh the mass of the combustion products that were released into the air and record the mass as M4.

Step 6: Compare the calculated mass of the combustion products in Step 4 to the measured mass of the combustion products in Step 5. If the mass of the combustion products released into the air is equal to the calculated mass, then the law of conservation of mass has been upheld. If not, then it has been violated and the cause of the difference should be identified and examined.

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calculate the ph for each case in the titration of 50.0 ml of 0.110 m hclo(aq) with 0.110 m koh(aq). use the ionization constant for hclo. what is the ph before addition of any koh?

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To calculate the pH at various points during the titration of 50.0 ml of 0.110 M HClO (aq) with 0.110 M KOH (aq), we need to determine the concentration of HClO and the concentration of OH- at each point. By using the ionization constant (Ka) for HClO, we can calculate the concentration of H3O+ (or H+) at the initial point before adding any KOH. The pH is then determined by taking the negative logarithm (base 10) of the H3O+ concentration.

HClO is a weak acid, and its ionization in water can be described by the following equation:

HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)

The ionization constant (Ka) for HClO is given as:

Ka = [H3O+][ClO-] / [HClO]

At the initial point before adding any KOH, the concentration of HClO is 0.110 M. Since no OH- has been added yet, the concentration of H3O+ is equal to the initial concentration of HClO.

Therefore, the pH at the initial point is determined by taking the negative logarithm (base 10) of the H3O+ concentration:

pH = -log[H3O+]

Substituting the concentration of H3O+, the pH can be calculated.

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determine the volume of 0.255 m koh solution required to neutralize each sample of sulfuric acid. the neutralization reaction is: h2so4(aq) 2koh(aq)→ k2so4(aq) 2h2o(l)

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The volume of 0.255 M KOH solution required to neutralize a sample of sulfuric acid can be determined by stoichiometry and the balanced equation of the neutralization reaction.

In the balanced equation for the neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH), it is stated that 2 moles of KOH react with 1 mole of H2SO4 to produce 2 moles of water (H2O) and 1 mole of potassium sulfate (K2SO4). This means that the stoichiometric ratio between H2SO4 and KOH is 1:2.

To determine the volume of 0.255 M KOH solution required, you need the molarity and volume of the sulfuric acid solution. Let's assume you have the volume of sulfuric acid in liters (L).

According to the stoichiometry, 1 mole of H2SO4 reacts with 2 moles of KOH. Using the molarity and volume of KOH, you can calculate the number of moles of KOH required. Then, based on the stoichiometric ratio, you can determine the moles of H2SO4 present in the sample.

Finally, using the molarity of the sulfuric acid solution, you can calculate the volume of the solution required to neutralize the given sample. Remember to convert the volume from liters to the desired unit (e.g., milliliters) if needed.

Please provide the volume of the sulfuric acid solution in order to proceed with the calculation.

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Calculate the pH and the equilibrium concentrations of HCO3- and CO32- in a 0.0514 M carbonic acid solution, H2CO3 (aq). For H2CO3, Ka1 = 4.2×10-7 and Ka2 = 4.8×10-11 pH = [HCO3-] = M [CO32-] = M

Answers

The pH of the carbonic acid solution is approximately 3.833, and the equilibrium concentrations of [tex][HCO_3^-][/tex] and [tex][CO_3^{2-}][/tex] are approximately 1.468 × [tex]10^{(-4)[/tex] M.

To calculate the pH and equilibrium concentrations of [tex][HCO_3^-][/tex] and [tex][CO_3^{2-}][/tex] in a carbonic acid solution, we need to consider the ionization reactions of carbonic acid [tex](H_2CO_3)[/tex]

The ionization reactions of carbonic acid are as follows:

[tex]H_2CO_3[/tex] ⇌ [tex]H^+[/tex] + [tex]HCO_3^-[/tex]

[tex]HCO_3^-[/tex] ⇌ [tex]H^+[/tex] + [tex]CO_3^{2-}[/tex]

Given:

Initial concentration of [tex]H_2CO_3[/tex] (carbonic acid): [tex][H_2CO_3][/tex] = 0.0514 M

Ka1 = 4.2 × [tex]10^{(-7)[/tex]

Ka2 = 4.8 × [tex]10^{(-11)[/tex]

[HCO3-] = M (equilibrium concentration)

[CO32-] = M (equilibrium concentration)

Step 1: Write the equilibrium expressions for the ionization reactions.

Ka1 = [tex][H^+][HCO_3^-]/[H_2CO_3][/tex]

Ka2 = [tex][H^+][CO_3^{2-}]/[HCO_3^-][/tex]

Step 2: Set up an ICE table (Initial, Change, Equilibrium) for each ionization reaction.

For reaction 1: [tex]H_2CO_3[/tex] ⇌ [tex]H^+ + HCO_3^-[/tex]

Initial: [tex][H_2CO_3][/tex] = 0.0514 M, [tex][H^+][/tex] = 0 M, [tex][HCO_3^-][/tex] = 0 M

Change: -x, +x, +x

Equilibrium: [tex][H_2CO_3][/tex] - x, x, x

For reaction 2: [tex]HCO_3^-[/tex] ⇌ [tex]H^+ + CO_3^{2-[/tex]

Initial: [tex][HCO_3^-][/tex] = 0 M, [tex][H^+][/tex] = 0 M, [tex][CO_3^{2-}][/tex] = 0 M

Change: +x, +x, +x

Equilibrium: [tex][HCO_3^-][/tex] + x, x, x

Step 3: Substitute the equilibrium concentrations into the equilibrium expressions and solve for x.

For reaction 1:

Ka1 = [tex][H^+][HCO_3^-]/[H_2CO_3][/tex]

4.2 × [tex]10^{(-7)[/tex] = x * x / (0.0514 - x)

Since the value of x is expected to be small compared to 0.0514, we can assume that (0.0514 - x) = 0.0514.

4.2 × [tex]10^{(-7)[/tex] = [tex]x^2[/tex] / 0.0514

Solving for x:

[tex]x^2[/tex] = 4.2 × [tex]10^{(-7)[/tex] * 0.0514

[tex]x^2[/tex] = 2.1588 × [tex]10^{(-8)[/tex]

x = 1.468 × [tex]10^{(-4)[/tex] M

Step 4: Calculate the pH.

The pH is determined by the concentration of [tex][H^+][/tex] ions. Since [tex][H^+][/tex] = x, the pH is equal to the negative logarithm of x.

pH = -log(x)

pH = -log(1.468 × [tex]10^{(-4)[/tex])

pH = 3.833

Step 5: Calculate the equilibrium concentrations of [tex][HCO_3^-][/tex] and [tex][CO_3^{2-}][/tex].

[tex][HCO_3^-][/tex] = [tex][H^+][/tex] = x

[tex][HCO_3^-][/tex] = 1.468 × [tex]10^{(-4)[/tex] M

[tex][CO_3^{2-}][/tex] = [tex][H^+][/tex] = x

[tex][CO_3^{2-}][/tex] = 1.468 × [tex]10^{(-4)[/tex] M

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does the strength of an acid and base impact the heat evolved by a neutralization reaction

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The strength of an acid and base does have an impact on the heat evolved by a neutralization reaction. Stronger acids and bases tend to produce more heat during neutralization compared to weaker acids and bases.

The heat evolved during a neutralization reaction is a result of the exothermic nature of the reaction, where energy is released. The strength of an acid or base is determined by its ability to donate or accept protons (H+) during the reaction. Strong acids and bases dissociate completely in water, releasing a higher concentration of H+ or OH- ions, respectively. When a strong acid reacts with a strong base, a larger number of H+ and OH- ions are available for neutralization, leading to a higher heat release.

In contrast, weak acids and bases only partially dissociate in water, resulting in a lower concentration of H+ or OH- ions. Consequently, when a weak acid reacts with a weak base, fewer H+ and OH- ions are available for neutralization, resulting in a lower heat release.

Therefore, the strength of an acid and base directly influences the concentration of H+ and OH- ions available for neutralization, ultimately impacting the heat evolved during the reaction. Stronger acids and bases produce a greater amount of heat, while weaker acids and bases result in a lower heat release.

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In a Zn/Cu cell, the standard cell potential is 1.10 V. How could you increase the voltage by changing the solution concentrations o f Zn²⁺ and Cu²⁺? Explain in words.

Answers

To increase the voltage in a Zn/Cu cell, you can change the solution concentrations of Zn²⁺ and Cu²⁺. By increasing the concentration of Zn²⁺ and decreasing the concentration of Cu²⁺, the voltage of the cell can be increased.

In a Zn/Cu cell, the standard cell potential of 1.10 V is determined by the difference in the standard reduction potentials of Zn²⁺ and Cu²⁺. By altering the concentrations of the ions, you can affect the reaction rates and shift the equilibrium of the cell reaction.

Increasing the concentration of Zn²⁺ increases the rate of the Zn oxidation reaction at the anode, while decreasing the concentration of Cu²⁺ decreases the rate of the Cu reduction reaction at the cathode. This leads to an increase in the overall voltage of the cell. The concentration changes affect the reaction rates, which in turn affect the flow of electrons and the overall voltage generated by the cell.

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Which of the following reactions is correctly balanced?
a. CO + O2 → CO2
b. N2 + H → 2NH3
c. Zn + 2HCl → H2 + ZnCl2
d. 2H2O + C → CO + 2H2

Answers

The number of carbon atoms, hydrogen atoms, and oxygen atoms on both sides of the equation is the same. Therefore, this equation is correctly balanced.

Reaction which one is correctly balanced is option (d) 2H2O + C → CO + 2H2. Explanation:In chemistry, balancing an equation is a process of changing coefficients to make both sides of a chemical equation equal. An equation that is balanced represents the laws of conservation of matter since it shows that the same number of atoms of each element is present on both sides of the equation. 1. CO + O2 → CO2CO is in the right proportion, but there are 3 oxygen molecules on the right and only 2 on the left, so this equation is unbalanced.2. N2 + H → 2NH3N is in the right proportion, but there are 3 hydrogen molecules on the right and only 1 on the left, so this equation is unbalanced.3. Zn + 2HCl → H2 + ZnCl2There is only one zinc on the left and one on the right, which is acceptable. However, there are only two chlorines on the left and two on the right, which is incorrect. Therefore, this equation is unbalanced.4. 2H2O + C → CO + 2H2In this equation, the number of carbon atoms, hydrogen atoms, and oxygen atoms on both sides of the equation is the same. Therefore, this equation is correctly balanced.

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Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements.
a. Rb and Cl
b. S and S
c. C and F
d. Ba and S
e. N and P
f. B and H

Answers

Rb and Cl, S and S, C and F, Ba and S, N and P, B and H form bond which is covalent.

Ionic, covalent, polar covalent bonds and the respective reasons:Ionics or ionic bond results due to the attraction between the opposite charges that arise from the transfer of one or more electrons from one element to the other element. Electronegativity difference ≥1.7.The bond formed between Rb and Cl is ionic because Rb (Rubidium) belongs to group 1A with one valence electron and Cl (Chlorine) belongs to group 7A with seven valence electrons. The electronegativity difference between them is 2.0, which is greater than the critical value of 1.7. Hence, RbCl is expected to be ionic.The bond formed between Sulfur and Sulfur is Covalent. It is due to the sharing of electrons between atoms of the same element. Electronegativity difference = 0.The bond formed between Carbon and Fluorine is Polar covalent. Carbon belongs to group 4A, and it has four valence electrons while fluorine belongs to group 7A and has seven valence electrons. The electronegativity difference between Carbon and Fluorine is 1.5. Therefore, the bond between Carbon and Fluorine is polar covalent.The bond formed between Barium and Sulfur is ionic. Ba belongs to group 2A, and S belongs to group 6A. The electronegativity difference between Ba and S is 2.6, which is greater than 1.7. Hence BaS is ionic.The bond formed between Nitrogen and Phosphorous is covalent. The reason is that they both belong to the same group (5A), and the electronegativity difference is also zero.The bond formed between Boron and Hydrogen is covalent. The reason is that B (Boron) belongs to group 3A, and H (Hydrogen) belongs to group 1A. The difference in electronegativity is only 0.9. Hence the bond is covalent.

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How many molecules are contained in 5.8 moles of acetaminophen molecules?

Answers

To determine the number of molecules in 5.8 moles of acetaminophen, we can use Avogadro's number as a conversion factor. Avogadro's number (6.022 × 10^23) represents the number of molecules in one mole of a substance. By multiplying the number of moles by Avogadro's number, we can find the number of molecules.

Avogadro's number is a fundamental constant that relates the number of particles (atoms, molecules, or ions) to the amount of substance in moles. It is approximately 6.022 × 10^{23} particles per mole. In this case, we have 5.8 moles of acetaminophen. To find the number of molecules, we can use the following conversion:

Number of molecules = Number of moles × Avogadro's number

Substituting the given values:

Number of molecules = 5.8 moles × 6.022 × [tex]10^{23}[/tex] molecules/mole

Calculating the result:

Number of molecules = 3.49556 × [tex]10^{24}[/tex] molecules

Therefore, there are approximately 3.49556 × 10^24 molecules in 5.8 moles of acetaminophen.

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Among these types of nucleons (odd and even numbers), which has the fewest stable nuclides?
A. odd number of protons and even number of neutrons o B. odd number of protons and odd number of neutrons o C.even number of protons and even number of neutrons o D. even number of protons and odd number of neutrons E. Odd or even numbers of nucleons does not influence the stability of nuclides QUESTION 3 F-17 undergoes positron decay. What is the product nucleus? Enter your answer using the same format, i.e, symbol-mass number

Answers

The answer is 17O-17, which is the product nucleus.

Among these types of nucleons, odd and even numbers of protons and neutrons, odd number of protons and even number of neutrons have the fewest stable nuclides. Let's see why:Odd number of protons and even number of neutrons has the fewest stable nuclidesOdd number of protons and even number of neutrons is an unstable combination because of the proton-neutron interactions, which results in an unequal distribution of nuclear force in the nucleus. In other words, this arrangement can lead to a destabilizing force, making it difficult for the nucleus to remain stable.Hence, among the given options, the answer is (A) odd number of protons and even number of neutrons.Now, let's move on to the next question.Question 3: F-17 undergoes positron decay. What is the product nucleus?The equation for the given nuclear reaction is: 9 17F → 8 17O + 1 0ePositron decay involves the conversion of a proton to a neutron, which can be represented by beta-plus emission. In this case, 17F (which contains nine protons) is transformed into 17O, which has eight protons, and a positron (0e or beta-plus particle). Thus, the answer is 17O-17, which is the product nucleus.

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write the equilibrium expression for the solubility of sodium cholride in water

Answers

The equilibrium expression for the solubility of sodium chloride (NaCl) in water is [Na⁺][Cl⁻].

The equilibrium expression for the solubility of sodium chloride (NaCl) in water is given by the equation NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq).

This equation represents the dissolution of solid sodium chloride in water, where the solid dissociates into individual sodium ions (Na⁺) and chloride ions (Cl⁻) in the aqueous phase.

The equilibrium expression can be written as [Na⁺] [Cl⁻], where [Na⁺] represents the concentration of sodium ions and [Cl⁻] represents the concentration of chloride ions in the solution at equilibrium.

The equilibrium constant (K) for this solubility equilibrium is the ratio of the concentrations of the dissociated ions to the concentration of the undissolved solid, and it provides a measure of the extent of dissolution of sodium chloride in water.

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The complete question is:

1. Write the equilibrium expression for the solubility of sodium chloride in water.

2. Describe what is happening at the molecular level in the saturated sodium chloride solution.

Which action would shift this reaction away from solid lead chloride and toward the dissolved ions?
O adding lead ions O adding chloride ions
O removing chloride ions
O removing lead chloride

Answers

The action that would shift this reaction away from solid lead chloride and toward the dissolved ions is by adding chloride ions.Why does adding chloride ions shift the reaction away from solid lead chloride and toward the dissolved ions?In a chemical reaction, the position of the equilibrium will shift when a stress is applied.

The stress may take the form of a concentration change, pressure change, or temperature change. Le Chatelier's Principle explains how a system at equilibrium responds to a stress. If the concentration of chloride ions is increased, the equilibrium will shift to the left, favoring the formation of more Pb2+ and Cl- ions to counteract the increase in chloride ions. If chloride ions are added to the solution containing solid lead chloride, the lead chloride will dissolve in the water to form dissolved lead and chloride ions, thus shifting the equilibrium toward the formation of dissolved ions.An increase in the concentration of lead ions would cause the equilibrium to shift toward the formation of solid lead chloride, which would be the opposite of what the question is asking. Therefore, the correct answer is adding chloride ions.

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which of the following best describes the interaction between water molecules?

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The interaction between water molecules is primarily characterized by hydrogen bonding.

Are water molecules connected by hydrogen bonds?

Water molecules exhibit a unique type of interaction known as hydrogen bonding. Hydrogen bonding occurs when the positively charged hydrogen atom of one water molecule is attracted to the negatively charged oxygen atom of another water molecule.

This creates a relatively strong intermolecular force that holds the water molecules together. In a water molecule, oxygen (O) and hydrogen (H) atoms are covalently bonded.

The oxygen atom has a slightly negative charge due to its higher electronegativity, while the hydrogen atoms carry a partial positive charge. This uneven distribution of charge within the molecule creates polar characteristics, making water a polar molecule.

When water molecules come into proximity, the positive end of one molecule (hydrogen) attracts the negative end of another molecule (oxygen). This attraction forms a hydrogen bond, which is a type of dipole-dipole interaction.

The hydrogen bond is weaker than a covalent or ionic bond but stronger than other intermolecular forces. Hydrogen bonding gives water several unique properties, including its high boiling and melting points, surface tension, and ability to dissolve a wide range of substances.

These properties are crucial for life on Earth as they facilitate various biological processes and allow water to act as a universal solvent. The unique properties of water and the role of hydrogen bonding in shaping its behavior are essential topics in chemistry and biology.

Understanding the nature of water molecules and their interactions provides insights into many scientific phenomena. Exploring the concept of hydrogen bonding in more depth can lead to fascinating discoveries in fields such as materials science, biochemistry, and environmental studies.

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the following equation shows the formation of water from hydrogen and oxygen. 2h2 o2 → 2h2o how many grams of water will form if 10.54 g h2 reacts with 95.10 g o2? g h2o

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The balanced chemical equation for the formation of water from hydrogen and oxygen is given below:

2H2(g) + O2(g) → 2H2O(l)

The equation shows that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

The molar mass of hydrogen is 2.016 g/mol, while that of oxygen is 32.00 g/mol.

Therefore, the number of moles of hydrogen that reacts can be determined as follows:

n(H2) = mass/Mr(H2)n(H2) = 10.54 g/2.016 g/moln(H2) = 5.23 mol

Similarly, the number of moles of oxygen can be calculated as follows:

n(O2) = mass/Mr(O2)n(O2) = 95.10 g/32.00 g/moln(O2) = 2.97 mol

From the balanced chemical equation, it can be seen that 2 moles of water is produced for every 2 moles of hydrogen and 1 mole of oxygen that react.

Therefore, the number of moles of water that is produced can be calculated as follows:

n(H2O) = 2 x n(O2)n(H2O) = 2 x 2.97n(H2O) = 5.94 mol

The mass of water produced can be determined using the following formula:

mass = n(H2O) x Mr(H2O)

mass = 5.94 mol x 18.015 g/mol

mass = 106.97 g

Thus, 106.97 g of water will be formed if 10.54 g H2 reacts with 95.10 g O2.

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Answer: its 94.03

Explanation: hydrogen

you have 19.7 grams of material and wonder how many moles were formed. your friend tells you to multiply the mass by grams/mole. is your friend correct?

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My friend is wrong because to calculate moles, we must divide the mass by molar mass.

How to calculate number of moles?

The number of moles of a substance refers to the base unit of amount of substance; the amount of substance of a system which contains exactly 6.022 × 10²³ elementary entities.

The number of moles in a substance can be calculated by dividing the mass of the substance by its molar mass as follows;

moles = mass (g) ÷ molar mass (g/mol)

According to this question, I have 19.7grams of a material and want to determine the number of moles in this material. The friend is wrong because moles can only be determined using the above method.

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1. Draw the transition state for the reaction of 1-chlorobutane and sodium hydroxide. 2. Explain why the following reaction results in racemization of the substitution product. CI CH3OH OCH3 OCH +

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The reaction of 2-chlorobutanol with methanol, leading to the formation of 2-methoxybutane, results in racemization due to the formation of a planar carbocation intermediate.

The reaction between 1-chlorobutanol and sodium hydroxide involves the substitution of chlorine (Cl) by hydroxide (OH) group, resulting in the formation of 1-butanol.

The transition state for this reaction can be envisioned as an intermediate state where the chlorine atom is partially dissociated from the carbon atom it is bonded to, while the hydroxide ion is approaching and getting closer to that carbon atom. In this transition state, the carbon atom undergoes hybridization changes, forming new bonds. The chlorine-carbon bond is weakened, and the carbon-oxygen bond is formed.

2. The reaction of 2-chlorobutanol with methanol to yield 2-methoxybutane involves the substitution of chlorine (Cl) by a methoxy group (OCH₃). This reaction leads to the racemization of the substitution product due to the involvement of an intermediate carbocation.

This reaction proceeds via a [tex]SN_1[/tex] (unimolecular nucleophilic substitution) mechanism, which involves the formation of a carbocation intermediate. During the formation of the carbocation intermediate, the chlorine atom dissociates from the carbon, leaving a positively charged carbon atom behind.

This intermediate carbocation is planar and lacks chirality. Consequently, when the methoxy group (OCH₃) attacks the carbocation from either face of the carbocation with equal likelihood, resulting in the formation of both enantiomers. This leads to a racemic mixture of the product, containing equal amounts of the R and S configurations.

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The complete question is:

1. Draw the transition state for the reaction of 1-chlorobutanol and sodium hydroxide. 2. Explain why the following reaction results in the racemization of the substitution product.

Reaction: 2-chlorobutanol gives rise to 2-methoxy butane in the presence of methanol.

Without doing any calculations, rank the following so that the least soluble is #1 and the most soluble is #5. v BaCO3, Ksp=5.0x10-9 v Cd(OH)2, Ksp=2.5x10-14 POCO3, Ksp=7.4x10-14 v AgCl, Ksp=1.8x10-10 v CaCrO4, Ksp=4.5x10-9

Answers

The solubility ranking of the compounds from least to most soluble is 1. AgCl, 2. CaCrO[tex]_4[/tex], 3. BaCO[tex]_3[/tex], 4. [tex]PO_4^{3-}[/tex], and 5. Cd(OH)[tex]_2[/tex].

It can be seen that all the salts are sparingly soluble. The solubility ranking of the compounds from least to most soluble is 1. AgCl, CaCrO[tex]_4[/tex], 3. BaCO[tex]_3[/tex], 4. [tex]PO_4^{3-}[/tex], 5. Cd(OH)[tex]_2[/tex].

Here, Ksp is the Solubility product constant. Ksp is the equilibrium constant for the dissolution of a sparingly soluble compound, meaning a compound that dissociates into a small number of ions in a solution.

The expression for the Ksp of a sparingly soluble salt is given as:

Ksp = [tex][A^{n+}][B^{m-}][/tex]

Where, A and B are the cation and anion of the sparingly soluble salt, and n and m are the corresponding stoichiometric coefficients of the cation and anion.

Thus, the order from least to most soluble is given by AgCl < CaCrO[tex]_4[/tex] < BaCO[tex]_3[/tex], < [tex]PO_4^{3-}[/tex] < Cd(OH)[tex]_2[/tex].

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in order to increase the amount of nh3 produced at equilibrium, do you need to increase or decrease the amount of n2 in the reactor? explain your reasoning.

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Yes, to increase the amount of NH[tex]_{3}[/tex] produced at equilibrium, you need to increase the amount of N[tex]_{2}[/tex] in the reactor.

According to Le Chatelier's principle, when a system at equilibrium is subjected to a change, it will shift in a direction that minimizes the effect of that change. In the reaction where NH[tex]_{3}[/tex] is produced from N[tex]_{2}[/tex] and H[tex]_{2}[/tex], increasing the amount of N[tex]_{2}[/tex] in the reactor will increase the concentration of N[tex]_{2}[/tex].

As a result, the equilibrium will shift in the forward direction to consume the excess N[tex]_{2}[/tex] and produce more NH[tex]_{3}[/tex] to restore equilibrium. Therefore, increasing the amount of N[tex]_{2}[/tex] in the reactor will favor the forward reaction and increase the amount of NH[tex]_{3}[/tex] produced at equilibrium.

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A student heated a 2.00 g of a hydrate of CoCl2 to drive off the water. After cooling the sample, the student found the mass of the anhydrate to be 1.09 g. 1. What is the mass (g) of water released? 2. What is the percent of water of hydration? 3. What is the formula of the hydrate?

Answers

The student heated a 2.00 g sample of a hydrate of CoCl2 and obtained 1.09 g of the anhydrate after cooling. The mass of water released can be calculated by subtracting the mass of the anhydrate from the initial mass of the hydrate: 2.00 g - 1.09 g = 0.91 g. Therefore, the mass of water released is 0.91 g.

To determine the percent of water of hydration, we need to compare the mass of water released to the mass of the original hydrate. The percent of water of hydration can be calculated using the formula: (mass of water released / mass of hydrate) x 100%. In this case, the mass of the original hydrate is 2.00 g. Plugging in the values, we get (0.91 g / 2.00 g) x 100% = 45.5%. Therefore, the percent of water of hydration is 45.5%.

To find the formula of the hydrate, we need to determine the ratio between the moles of water and the moles of anhydrate. We can use the molar masses of water (18.015 g/mol) and CoCl_{2} (129.841 g/mol) to calculate the moles of each component. The moles of water can be found by dividing the mass of water released by its molar mass: 0.91 g / 18.015 g/mol = 0.0505 mol. The moles of anhydrate can be calculated similarly: 1.09 g / 129.841 g/mol = 0.0084 mol. To find the ratio, we divide the moles of water by the moles of anhydrate: 0.0505 mol / 0.0084 mol ≈ 6.01. This means that the formula of the hydrate is[tex]CoCl_{2}[/tex]·[tex]6H_{2}O[/tex], indicating that for every[tex]CoCl_{2}[/tex] unit, there are six water molecules associated with it.

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Final answer:

The mass of water released is 0.91 g. The percent of water of hydration is 45.5%. The formula of the hydrate is likely CoCl2 · H2O.

Explanation:

To determine the mass of water released, we subtract the mass of the anhydrate from the mass of the initial hydrate: 2.00 g - 1.09 g = 0.91 g. The percent of water of hydration can be calculated by dividing the mass of water released by the mass of the initial hydrate and multiplying by 100: (0.91 g / 2.00 g) * 100 = 45.5%. From the information given, we can find the formula of the hydrate by comparing the molar masses of the anhydrate and the water: (1.09 g / 0.91 g) = 1.20. Since this is close to a 1:1 ratio, the formula of the hydrate is likely CoCl2 · H2O.

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Match the similarities between an electrical circuit and water circuit.
1. PSI A. battery
2. pipes B. open circuit
3. pump C. ammeter
4. valve D. switch
5. restriction E. amps
6. water meter F. resistance
7. water G. + voltage
8. high pressure output H. - voltage
9. low pressure intake I. closed circuit
10. valve closed J. electrons
11. valve open K. volts
12. liters/second L. conductors

Answers

In an electrical circuit, the similarities with a water circuit can be observed as follows:

PSI (Pounds per Square Inch) in water circuit corresponds to voltage in an electrical circuit. Both represent the potential energy or pressure that drives the flow.Pipes in a water circuit are equivalent to conductors in an electrical circuit. They provide a path for the fluid or electricity to flow through.Pump in a water circuit is analogous to a battery in an electrical circuit. They both act as the source of energy, providing the initial push or force to move the fluid or electrons.Valve in a water circuit is similar to a switch in an electrical circuit. Both can control or regulate the flow, either by opening or closing the circuit/pathway.Restriction in a water circuit can be compared to resistance in an electrical circuit. They both impede the flow and reduce the overall current or flow rate.Water meter in a water circuit corresponds to an ammeter in an electrical circuit. They measure the flow rate or current passing through the circuit.Water itself in a water circuit is akin to electrons in an electrical circuit. Both act as the carriers of energy, either in the form of fluid or electric charge.High pressure output in a water circuit is equivalent to positive voltage in an electrical circuit. It represents the part of the circuit with higher potential energy or pressure.Low pressure intake in a water circuit is analogous to negative voltage in an electrical circuit. It represents the part of the circuit with lower potential energy or pressure.

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Consider the reaction
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
How much Na3PO4 can be prepared by the
reaction of 6.3 g of H3PO4 with an excess of
NaOH?

Answers

The amount of Na3PO4 that can be prepared by the reaction of 6.3 g of H3PO4 with an excess of NaOH is 10.496 g.

The balanced chemical reaction isH3PO4 + 3 NaOH → Na3PO4 + 3 H2OHere we are given that 6.3 g of H3PO4 reacts with an excess of NaOH to form Na3PO4. We need to calculate how much Na3PO4 is formed.The molecular weight of H3PO4 is 98 g/mol. Thus, the number of moles of H3PO4 is given as= 6.3 g/ 98 g/mol= 0.064 moles The balanced chemical reaction tells us that 1 mole of H3PO4 reacts with 3 moles of NaOH to form 1 mole of Na3PO4. Thus, the number of moles of Na3PO4 formed is given by= 0.064 moles H3PO4 × 1 mole Na3PO4 / 1 mole H3PO4= 0.064 moles Na3PO4Therefore, the number of grams of Na3PO4 formed is given by= number of moles × molecular weight= 0.064 moles × 164 g/mol= 10.496 g

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Blasting caps containing Lead Azide detonate. How fast is the chemical reaction occurring? Subsonic speeds (slower than the speed of sound) Supersonic speeds (faster than the speed of sound) Submit Activate Win

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The chemical reaction in blasting caps containing Lead Azide occurs at supersonic speeds, indicating a highly explosive and rapid reaction. The release of energy is sudden and violent, leading to the detonation of the blasting caps.

The detonation of blasting caps containing Lead Azide can occur at both subsonic and supersonic speeds, depending on the conditions and the specific characteristics of the reaction.

At subsonic speeds, the chemical reaction occurs relatively slowly compared to the speed of sound. The reaction proceeds through a series of chemical reactions and propagation of shockwaves within the blasting cap. The reaction front moves at speeds lower than the speed of sound, resulting in a relatively slower detonation process.

On the other hand, at supersonic speeds, the chemical reaction occurs rapidly, faster than the speed of sound. The detonation process involves a high-speed shockwave that propagates through the blasting cap, causing almost instantaneous chemical reactions and energy release. This supersonic detonation can result in a more rapid and violent explosion.

The specific speed at which the chemical reaction occurs in Lead Azide blasting caps depends on various factors such as the initiation mechanism, the confinement conditions, and the specific composition of the explosive material.

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calculate the ph of 0.00345 m solution of strontium hydroxide

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The pH of the 0.00345 M solution of strontium hydroxide will be approximately 11.84.

Strontium hydroxide (Sr(OH)₂) is the strong base which dissociates completely in water. To calculate the pH of a 0.00345 M solution of strontium hydroxide, we need to determine the concentration of hydroxide ions (OH⁻) in the solution.

Since strontium hydroxide dissociates into two hydroxide ions per formula unit, the concentration of hydroxide ions (OH⁻) will be twice the concentration of strontium hydroxide.

Concentration of OH- = 2 × 0.00345 M = 0.0069 M

To calculate the pOH of the solution, we can use the formula:

pOH = -log10[OH⁻]

pOH = -log10(0.0069) ≈ 2.16

Finally, to calculate the pH of the solution, we will use the relationship;

pH + pOH = 14

pH + 2.16 = 14

pH ≈ 14 - 2.16 ≈ 11.84

Therefore, the pH of strontium hydroxide is approximately 11.84.

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determine the number of electron groups around the central atom for each of the following molecules.

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CH4, NH3, and H2O all have four electron groups around their respective central atoms.

In chemistry, the electron group refers to the number of electron pairs around the central atom of a molecule. This value helps determine the shape and structure of the molecule, as well as its properties. In this question, we are asked to determine the number of electron groups around the central atom for each of the given molecules.

Firstly, we have to determine the central atom in each molecule. This is usually the least electronegative element in the compound, which is often the one that appears less frequently. The central atom is connected to other atoms in the molecule through covalent bonds.

In the first molecule, CH4, the central atom is carbon. Carbon is bonded to four hydrogen atoms, each of which contributes one electron to a shared pair. Therefore, there are four electron pairs around the central carbon atom, which means there are four electron groups in CH4.

In the second molecule, NH3, the central atom is nitrogen. Nitrogen is bonded to three hydrogen atoms, each of which contributes one electron to a shared pair. In addition, nitrogen has a lone pair of electrons. Therefore, there are four electron pairs around the central nitrogen atom, which means there are four electron groups in NH3.

In the third molecule, H2O, the central atom is oxygen. Oxygen is bonded to two hydrogen atoms, each of which contributes one electron to a shared pair. In addition, oxygen has two lone pairs of electrons. Therefore, there are four electron pairs around the central oxygen atom, which means there are four electron groups in H2O.

In conclusion, CH4, NH3, and H2O all have four electron groups around their respective central atoms.

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