2 questions. will give brainliest if i can figure out how

What happens to the electrons of a reducing agent during a reaction?
a.Electrons are gained
b.Electrons are lost
c.Electrons are held
d.Electrons are doubled

When is a substance reduced during an oxidation reduction reaction?
a.After one substance is oxidized
b.Before one substance is oxidized
c.Same time as one substance is being oxidized
d.Only oxidation occurs

The amphiprotic substances among the options provided are HS⁻ and NH₃.

The term "amphiprotic" refers to substances that can act as both acids and bases, depending on the reaction conditions.

a. H₂SO₄ (sulfuric acid) is not considered amphiprotic. It is a strong acid that readily donates a proton (H+) but does not accept protons in typical acid-base reactions.

b. HS⁻ (hydrogen sulfide ion) is amphiprotic. It can act as a base by accepting a proton to form H₂S (hydrogen sulfide), and it can also act as an acid by donating a proton to form S²⁻ (sulfide ion).

c. PO₄³⁻ (phosphate ion) is not considered amphiprotic. It is a polyprotic base that can donate multiple protons, but it does not accept protons in typical acid-base reactions.

d. NH₃ (ammonia) is amphiprotic. It acts as a base by accepting a proton to form NH₄⁺ (ammonium ion), and it can also act as an acid by donating a proton to form NH₂⁻ (amide ion).

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Answer:

producer to decomposer

Explanation:

This is because in a food chain , energy flow from one trophic level to another. The producer which include plants are the source of energy which they manufacture good in the presence of light energy from sun. Energy flow directly from the producer to the primary consumer which are heterotrophs that feed on plants. Energy flow from consumer to decomposer after the consumer died and it is decayed.

CBr₄ has a higher vapor pressure at the same temperature than CCl₄. The statement is False.

The vapor pressure of a liquid is the pressure exerted by the vapor of the liquid in equilibrium with the liquid. The vapor pressure of a liquid depends on the strength of the intermolecular forces in the liquid. The stronger the intermolecular forces, the lower the vapor pressure.

In the case of CCl₄ and CBr₄, the intermolecular forces are London dispersion forces. The size of the atom or molecule affects the strength of the London dispersion forces. The larger the atom or molecule, the stronger the London dispersion forces.

Bromine is larger than chlorine, so the intermolecular forces in CBr₄ are stronger than the intermolecular forces in CCl₄. Therefore, CCl₄ will have a higher vapor pressure than CBr₄ at the same temperature.

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To calculate the mass in grams of 8.35 × 10²² molecules of CBr₄ (carbon tetrabromide), we need to use Avogadro's number to convert the given number of molecules to moles and then use the molar mass of CBr₄ to convert moles to grams.

The molar mass of CBr₄ can be calculated by adding up the atomic masses of carbon (C) and four bromine (Br) atoms. The atomic mass of carbon is approximately 12.01 g/mol, and the atomic mass of bromine is approximately 79.90 g/mol.

Molar mass of CBr₄ = (1 × 12.01 g/mol) + (4 × 79.90 g/mol) = 331.74 g/mol

To convert the number of molecules to moles, we divide the given number of molecules by Avogadro's number (6.022 × 10²³ molecules/mol):

Moles of CBr₄ = (8.35 × 10²² molecules) / (6.022 × 10²³ molecules/mol) = 0.138 mol

Finally, to find the mass in grams, we multiply the number of moles by the molar mass:

Mass of CBr₄ = (0.138 mol) × (331.74 g/mol) = 45.80 g

Therefore, the mass in grams of 8.35 × 10²² molecules of CBr₄ is approximately 45.80 grams.

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Answer:

4. Reactant

5. Left

I hope this helps ❤️❤️

Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.

Option (c) is correct.

To determine if the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL based on the given concentration of OH⁻, we need to calculate the solubility of Ca(OH)₂ using the provided concentration.

The balanced chemical equation for the dissociation of Ca(OH)₂ is:

Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻

From the equation, we can see that for every mole of Ca(OH)₂ that dissolves, two moles of OH⁻ ions are produced. Therefore, the concentration of OH⁻ is twice the concentration of Ca(OH)₂.

Given that the concentration of OH⁻ is 5.2 x 10⁻² M, the concentration of Ca(OH)₂ can be calculated by dividing the concentration of OH⁻ by 2:

Ca(OH)₂ concentration = (5.2 x 10⁻² M) / 2 = 2.6 x 10⁻² M

To determine the solubility of Ca(OH)₂ in grams per 100 mL, we can use the molar mass of Ca(OH)₂:

Solubility of Ca(OH)₂ = (2.6 x 10⁻² M) x (74.1 g/mol) = 1.924 g/100 mL

Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.

Therefore, the correct answer is c) Yes.

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Answer:

Explanation:

true.                        

Answer:

True. why?

Because the fluid that is lost through the blood vessels is restored but with no central pump or heart, this device releases and collects fluid.

Answer:

Because both the bonding electrons come from the oxygen atom. Explanation: A coordinate covalent bond is formed when both the bonding electrons are coming from the same atom

Explanation:

The addition of 0.183 moles of hydrobromic acid to a 1 liter solution containing 0.487 M acetic acid and 0.365 M sodium acetate will lower the pH slightly. On the other hand, the addition of 0.391 moles of nitric acid to a 1 liter solution containing 0.474 M hydrocyanic acid and 0.356 M potassium cyanide will lower the pH by several units.

In the first case, the acetic acid and sodium acetate form a buffer solution, which helps maintain a relatively stable pH. The acetic acid acts as a weak acid, while the sodium acetate is its conjugate base. When a small amount of hydrobromic acid is added, it reacts with the acetate ions in the buffer system, shifting the equilibrium towards the formation of more acetic acid and water. This results in an increase in the concentration of hydronium ions (H3O+), leading to a slight decrease in pH.

In the second case, the hydrocyanic acid and potassium cyanide also form a buffer solution. However, nitric acid is a strong acid that completely ionizes in water. The addition of nitric acid increases the concentration of hydronium ions significantly, overpowering the buffer system and causing a substantial decrease in pH.

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Answer:

A

Explanation:

f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.

Hence, the correct option is C.

Electronegativity is a measure of an atom's ability to attract electrons towards itself when it is part of a chemical bond. It generally increases as you move across a period from left to right and decreases as you move down a group in the periodic table.

A. cl < f < se < as (chlorine, fluorine, selenium, arsenic):

Chlorine (Cl) has a higher electronegativity than fluorine (F), selenium (Se), and arsenic (As). Therefore, it is arranged correctly that chlorine has the highest electronegativity, followed by fluorine, selenium, and then arsenic.

B. se < f < as < cl (selenium, fluorine, arsenic, chlorine):

This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then arsenic (As), and selenium (Se) has the lowest electronegativity.

C. f < cl < se < as (fluorine, chlorine, selenium, arsenic):

This arrangement is correct. Fluorine (F) has the highest electronegativity, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.

D. as < se < cl < f (arsenic, selenium, chlorine, fluorine):

This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.

Therefore, the correct answer is option C: f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.

Hence, the correct option is C.

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The most appropriate explanation among the given options for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺ is that "Li⁺ has a smaller ionic radius than K⁺." Option D is correct.

Ionic radius refers to the size of an ion, specifically the distance between the nucleus and the outermost electron shell. Smaller ions tend to have stronger interactions with solvent molecules because the charge is more concentrated in a smaller space, leading to a higher electrostatic attraction between the ion and the surrounding solvent molecules.

In this case, Li⁺ has a smaller ionic radius compared to K⁺. As a result, Li⁺ has a stronger interaction with the solvent molecules, which leads to a greater ion-solvent interaction.

Hence, D. is the correct option.

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--The given question is incomplete, the complete question is

"Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺? A) The ionization energy of Li is higher than that for K. B) Li has a lower density than K. C) Li⁺ is of a lower mass than K⁺ D) Li⁺ has a smaller ionic radius than K⁺ E) Li reacts with water more slowly than K."--

Answer:

0.68 M

Explanation:

The concentration of a solute refers to Molarity= [tex]\frac{moles}{Liter}[/tex]

We need to find moles, we can do that by using stoichiometry to convert 6g > moles using molar mass.

Molar mass of C6H8O6= 176 g/mol; this is our key to converting the amount we have to moles.

6g C6H8O6 x [tex]\frac{1 mol}{176 g}[/tex]C6H8O6 = 0.034 mol C6H8O6

Now we need our volume per drop.

We know 1000 mL=1 L

We have 20 drops per 1 mL, so 1 mL / 20 = 0.05 mL per drop. But we need Liters, so we'll convert it to Liters.

0.05 mL x [tex]\frac{1 L}{1000mL}[/tex] = 0.00005L

Now we can solve for Molarity.

Molarity of C6H8O6 = [tex]\frac{0.034 mols}{0.00005 L}[/tex] = 0.068M of C6H8O6

Answer:

there will be winds moving from 25miles per hour to 30 miles per hour towards the east

The final concentration of the solution is determined as 0.59 M.

The concentration of a solution or molarity of a solution is defined as the ratio of number of moles of solute to the liters of the solution.

Molarity, M = n / V

where;

The final concentration of the solution is calculated as follows;

the number of moles = 0.5 m

the total volume after adding 500 ml of water = 340 ml + 500 ml = 840 ml = 0.84 L

Molarity = 0.5 m / 0.84 L

Molarity = 0.59 M

Thus, the final concentration of the solution is determined as 0.59 M.

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Answer: Balanced forces

Explanation: Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion.

Answer:

A. 1147.44 grams

Explanation:

Multiply volume by molarity to get the moles of solution.

5.85 M = mol/2.0 L

mol = (5.85 M)(2.0 L)

mol = 11.7

There is 11.7 moles of H2SO4.

Convert to grams with molar mass.

11.7 mol H2SO4 x (98.076 g/1 mol) = 1147.49 g

Closest answer is A, 1147.44 g.

To calculate the percent recovery for each component (basic, acidic, and neutral) and the percent error for the melting point of each component in Experiment 4C, specific formulas and calculations are required. The percent recovery is calculated by dividing the mass of the recovered component by the initial mass and multiplying by 100. The percent error for the melting point is calculated by comparing the experimental melting point to the accepted literature value and expressing the difference as a percentage of the accepted value.

To calculate the percent recovery for each component, you need to divide the mass of the recovered component by the initial mass and multiply by 100. Let's perform the calculations for each component:

a. Basic (Ethyl-4-amino benzoate):

Percent Recovery = (Recovered mass of basic component / Initial mass of basic component) x 100

Percent Recovery = (0.0137 g / 0.0498 g) x 100 = 27.51%

b. Acidic (Benzoic Acid):

Percent Recovery = (Recovered mass of acidic component / Initial mass of acidic component) x 100

Percent Recovery = (0.0322 g / 0.0588 g) x 100 = 54.76%

c. Neutral (9-fluorenone):

Percent Recovery = (Recovered mass of neutral component / Initial mass of neutral component) x 100

Percent Recovery = (0.0422 g / 0.0508 g) x 100 = 83.07%

To calculate the percent error for the melting point of each component, you need to compare the experimental melting point to the accepted literature value. The percent error is calculated using the formula:

Percent Error = ((Experimental melting point - Accepted melting point) / Accepted melting point) x 100

Let's perform the calculations for each component:

a. Basic (Ethyl-4-amino benzoate):

Percent Error = ((84.0°C - Accepted melting point) / Accepted melting point) x 100

b. Acidic (Benzoic Acid):

Percent Error = ((121.5°C - Accepted melting point) / Accepted melting point) x 100

c. Neutral (9-fluorenone):

Percent Error = ((82.0°C - Accepted melting point) / Accepted melting point) x 100

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The balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.

Redox reaction involving the permanganate ion in acidic solution:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OThe balanced redox reaction that involves the permanganate ion in acidic solution is given by:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OFor balancing the given reaction in acidic solution, we follow the given steps:Step 1: Writing the unbalanced reaction in the ionic form:Step 2: Separating the half-reactions for oxidation and reduction:Oxidation half-reaction: 5NO2− → 5NO3−Reduction half-reaction: 2MnO4− → 2Mn2+Step 3: Balancing the number of atoms of elements in half-reactionsBalancing oxidation half-reaction:5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Step 4: Balancing the number of electrons lost and gained in each half-reaction by multiplying by a coefficient.Oxidation half-reaction: 5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Now, the electrons lost and gained are balanced.Step 5: Balancing the number of hydrogen ions in each half-reaction:Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Step 6: Balancing the number of oxygen atoms in each half-reaction by adding H2O molecules to the side that needs it. For acidic solution, add H+ ions to balance the oxygen atoms.Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Now, the number of oxygen atoms is balanced.Step 7: Adding half-reactions, multiplying by integers (if necessary) to equalize the electrons in each half-reaction, and canceling species common to both sides.5NO2− + 2MnO4− + 16H+ → 5NO3− + 2Mn2+ + 8H2OFinally, the balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.

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In the given energy chain, the energy is created in the fire when it heats up the pot. When the pot heats up, it cooks the food that fuels the body. As the food fuels the body, the body releases heat during activity. Therefore, the energy is created in the initial stage of the chain, i.e., when the fire heats up the pot.

The given energy chain is as follows:

Fire heats up a pot → Pot cooks food → Food fuels body → Body releases heat during activity

In the given energy chain, the energy is created in the first step when the fire heats up the pot. As the fire heats up the pot, the pot becomes hot and cooks the food that is in it. This cooking of food releases energy, which can be used by the body.The energy is then transferred to the body as it consumes the cooked food. The food fuels the body and provides it with the necessary energy to carry out activities. As the body uses this energy, it also releases heat during activity, which can be measured in the form of sweat or an increase in body temperature.Therefore, the energy is created in the fire heats up the pot step of the energy chain.

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Answer:

10.4 hours

Explanation:

Now from;

0.693/t1/2 = 2.303/t log (No/N)

t1/2= half life of the radioactive isotope

t = age of the radioactive isotope

No= Original amount of radioactive isotope

N= amount of radioactive isotope left after time t

Substituting values;

N = 0.0625No

0.693/2.6 = 2.303/t log (No/0.0625No)

0.693/2.6 = 2.303/t log (1/0.0625)

0.267 = 2.303/t * log 16

0.267 = 2.773/t

t = 2.773/0.267

t = 10.4 hours

Answer:

I only

last answer

Explanation:

Answer:

B. An equation that shows only those particles involved in the reaction, and that is balanced for both mass and charge

Explanation:

A net ionic equation is a chemical equation that will only show the elements, compounds, and ions that are directly involved in the chemical reaction. It comes from the balanced molecular equation and should be balanced by both mass and charge.

Answer:

moles Pb(NO₃)₂ ≅ 0.0213 mole (three sig. figs.)

Explanation:

converting...

grams to moles => divide by formula weight

moles to grams => multiply by formula weight

for this problem ...

moles Pb(NO₃)₂ = 7.04g / 331.2g/mol = 0.021256 (calculator answer)

≅ 0.0213 mole (three sig. figs.)

Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore,  the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is 0.0212moles.

The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.

Mathematically,

mole =given mass ÷ molar mass

Molar mass of 1 mole of  Pb(NO₃)₂ = 331.2g/mol

given mass of Pb(NO₃)₂=7.04g

Substituting the values in above formula we get

moles of Pb(NO₃)₂ = 7.04g / 331.2g/mol

moles of Pb(NO₃)₂ = 0.0212moles

Therefore,  the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is  0.0212moles.

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Part A: [tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) → [tex]Fe^3+(aq)[/tex]+ [tex]3H_{2}O(l)[/tex](solid [tex]Fe(OH)_{2}[/tex] reacts with aqueous H+ to form [tex]Fe^3+[/tex] ions and water). Part B: H+(aq) + [tex]ClO_{3}^-[/tex](aq) + Na+(aq) +[tex]OH^-(aq)[/tex]→ Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex](H+ reacts with [tex]OH^-[/tex] to form water in the presence of Na+ and [tex]ClO_{3}^-[/tex] ions).

Part A:

The balanced net ionic equation for the reaction between solid [tex]Fe(OH)_{3}[/tex] and aqueous [tex]H_{2}SO_{4}[/tex]is:

[tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) + [tex]3SO_{4}^2-(aq)[/tex] → [tex]Fe^3+(aq)[/tex] + [tex]3SO_{4}^2-(aq)[/tex] +[tex]3H_{2}O(l)[/tex].

In this reaction, the [tex]Fe(OH)_{3}[/tex] solid reacts with the H+ ions from [tex]H_{2}SO_{4}[/tex]to form[tex]Fe^3+[/tex] ions and water molecules. The sulfate ions [tex](SO_{4}^2-)[/tex]are spectator ions and do not participate in the net reaction. The phases are indicated by (s) for solid, (aq) for aqueous, and (l) for liquid.

Part B:

The balanced net ionic equation for the reaction between [tex]HClO_{3}(aq)[/tex] and NaOH(aq) is:

H+(aq) + [tex]ClO_{3}^-(aq)[/tex]+ Na+(aq) + [tex]OH^-(aq)[/tex] → Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex].

Here, the H+ ion from [tex]HClO_{3}[/tex] reacts with the [tex]OH^-[/tex] ion from NaOH to form water molecules. Sodium ion (Na+) and chlorate ion ([tex]ClO_{3}^-[/tex]) are spectator ions. The phases are indicated by (aq) for aqueous and (l) for liquid.

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Answer:  Shaun must consume 6 more g of protein to reach the required amount.

Explanation:

Given : Recommended amount of protein = 200 calories

Now 4 calories = 1 gram of protein

Thus 200 calories = [tex]\frac{1}{4}\times 200=50[/tex] g of protein.

Calories consumed per day by Shaun = 176 calories

176 calories =  [tex]\frac{1}{4}\times 176=44[/tex] g of protein.

Thus Shaun need to consume = (50-44) g = 6 of protein to reach the required amount.

Shaun must consume 6 more g of protein to reach the required amount.

Answer:

2 Mg + O2 -> 2 MgO

is a synthesis reaction.

The critical angle is given by sin ic = n2 / n1, where n1 and n2 are the indices of refraction for the media.

Using the formula sin ic = n2 / n1 for the critical angle for a liquid-air surface, we can calculate the index of refraction of the liquid.47.7° = sin⁡ic = n2 / n1, where n1 is the index of refraction of air, which is 1.00. The index of refraction of the liquid is therefore 1.33.

When light travels from a medium with a high refractive index (n1) to a medium with a low refractive index (n2), total internal reflection occurs if the angle of incidence is greater than the critical angle (ic).

The critical angle (ic) can be calculated using the formula sin ic = n2 / n1, where n1 and n2 are the indices of refraction of the media. The angle of incidence that causes total internal reflection is the critical angle.

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For the reaction 2H2S(g) ⇌ 2H2(g) + S2(g), the effect of a volume change on this reaction at equilibrium can be considered. It can be predicted whether increasing the reaction volume will cause the reaction to shift to the right, shift to the left, or have no effect on the reaction.

In this reaction, two moles of gaseous reactants are being converted into two moles of gaseous products. Therefore, this is a reaction that involves a change in volume. According to Le Chatelier's principle, the equilibrium will shift in such a way as to counteract the effect of the volume change.

If the reaction volume is increased by adding inert gas (i.e., a gas that doesn't react with the reactants), then the total pressure of the system will increase, but the partial pressures of the reactants and products will remain the same. Since the partial pressures remain unchanged, the equilibrium will not shift in either direction.

However, if the reaction volume is decreased by removing some of the gaseous products, then the partial pressure of the products will decrease. In order to counteract this decrease, the equilibrium will shift to the right to produce more gaseous products (i.e., more H2 and S2). Therefore, decreasing the volume will cause the reaction to shift to the right.

Therefore, increasing the reaction volume will have no effect on this reaction at equilibrium.

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