a 30.0- ml volume of 0.50 m ch3cooh ( ka=1.8×10−5 ) was titrated with 0.50 m naoh . calculate the ph after addition of 30.0 ml of naoh at 25 ∘c . express the ph numerically.

After considering the given data we conclude that the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K is [tex]4.8*10^{-17}[/tex].

To evaluate the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K, we can apply the following steps:
Write the balanced chemical equation for the reaction.
[tex]N_2(g) + 2NO_2(g) - - - > 2NO_2(g)[/tex]
Write the expression for Kp for each of the two reactions given.
[tex]Kp_1 = [NO]^2/[N_2][O_2][/tex]
[tex]Kp_2 = [NO_2]^2/[NO]^2[O_2][/tex]
Here,[tex][NO], [N_2], [O_2], and [NO_2][/tex]are the partial pressures of the respective species at equilibrium.
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex] into the expression for Kp for the desired reaction.
[tex]Kp = Kp_2/Kp_1 = ([NO_2]^2/[NO]^2[O_2])/([NO]^2/[N_2][O_2]) = ([NO_2]^2[N_2])/[NO]^4[/tex]
Restructure the expression for Kp to solve for the unknown partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex], and the given partial pressures of N2 and O2, into the expression for [NO].
[tex][NO] = \sqrt(Kp_1[N_2][O_2])[/tex]
Stage the calculated value of [NO] into the expression for[tex][NO_2][/tex] to obtain the equilibrium partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
[tex][NO_2]^2 = (2.4*10^{12} )(4.0*10^{-31} )([N_2][O_2])^2/[N_2][/tex]
[tex][NO_2]^2 = 9.6*10^{-19} [N_2][O_2]^2[/tex]
[tex][NO_2] = \sqrt(9.6*10^{-19} [N_2][O_2]^2)[/tex]
[tex][NO_2] = 3.1*10^{-10} atm[/tex]
Stage the evaluated values of [NO] and[tex][NO_2][/tex] into the expression for Kp to obtain the equilibrium constant.
[tex]Kp = ([NO_2]^2[N_2])/[NO]^4[/tex]
[tex]Kp = (9.6*10^{-19} )(3.1*10^{-10} )^2/[/\sqrt(4.0*10^{-31*1} atm)]^4[/tex]
[tex]Kp = 4.8*10^{-17}[/tex]
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The minimum hydroxide concentration required for the precipitation of [tex]Fe(OH)_2[/tex] to begin is [tex]1.691* 10^-^8 M.[/tex]

The balanced chemical equation for the precipitation reaction is:

[tex]Fe(OH)_2[/tex](s) ⇌ [tex]Fe^2^+[/tex](aq) + [tex]2OH^-[/tex](aq)

Ksp = [tex][Fe^2^+][OH^-]^2[/tex]

Ksp [tex]Fe(OH)_2[/tex] = [tex]4.87*10^-^1^7[/tex]

Ksp = [tex][Fe^2^+][OH^-]^2[/tex]

[tex]4.87*10^-^1^7[/tex] = (0.085)(2x)²

[tex]4.87*10^-^1^7[/tex] = 0.17x²

x² =[tex]4.87*10^-^1^7[/tex] / 0.17

x² = [tex]2.8647* 10^-^1^6[/tex]

x = [tex]1.691*10^-^8[/tex] M

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Answer:

CH4 and O2

Explanation:

The reactants of a chemical reaction are those that are on the left side of the chemical equation. In this case, methane (CH4) and oxygen (O2) which are on the left side, are the reactants.

You divide by the mass of the solution.

Percent by mass is the mass of the solute divided by the mass of the solution, multiplied by 100.

Example:

You dissolve a 4.00 g sugar cube in 350 mL of tea at 80 °C. The density of water at 80 °C is 0.975 g/mL. What is the percent by mass of sugar in the solution?

Solution

Step 1 — Determine mass of the solute.

The solute is the sugar cube. Mass of solute = 4.00 g.

Step 2 — Determine mass of solvent.

The solvent is the 80 °C water. Use the density of the water to find the mass.

mass of solvent = 350 mL ×

0.975

g

1

mL

= 341 g

Step 3— Determine the total mass of the solution

mass of solution = mass of solute + mass of solvent = 4.00 g + 341 g = 345 g

Step 4 — Determine percent composition by mass of the sugar solution.

% by mass =

mass of solute

mass of solution

× 100 % =

4.00

g

345

g

× 100 % = 1.159 %

The percent composition by mass of the sugar solution is 1.159 %.

Answer:

left is chemical because its chemicals, and right is physical because the ice cream is melting and thats physical

Explanation:

Barium nitrate and sodium chloride do not react with each other because both of them are soluble in water. The chemical equation for the reaction between barium nitrate and sodium chloride is given below.

Ba(NO₃)₂ + 2NaCl → BaCl₂ + 2NaNO₃

The reaction between barium nitrate and sodium chloride is a double displacement reaction, where barium cation is exchanged with sodium cation and nitrate anion is exchanged with chloride anion. But the reaction does not occur due to the solubility of barium nitrate and sodium chloride in water.The solution of barium nitrate and sodium chloride will remain clear and colorless with no precipitation forming. In fact, it is a method of testing the presence of sulfate ions in a solution. A small amount of barium nitrate is added to the solution to form barium sulfate. Since barium sulfate is insoluble, it forms a white precipitate.

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Water evaporates more slowly than alcohol because water has more surface tension and more cohesive force than alcohol, So it takes more time for water to evaporate than for alcohol.

In alcohol, only O-H present is hydrophilic i.e. polar. Apart from this alcohol has a large hydrocarbon group i.e. C-H present in it, which is non-polar in nature. As a result, the attractive force between the atoms in alcohol is less. As a result, during boiling, the bonds between the atoms vibrate and break quickly. So alcohol evaporates faster.

In water, the entire molecule is polar. So there exists a strong attractive force between them. So the cohesive force is more in the case of water as compared to that of alcohol. As water is boiled the O-H bonds start to vibrate, since the bond is stronger, it requires higher energy for boiling. Thus it evaporates slower.

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Answer:

True

Explanation:

DON'T GIVE BRAINLIEST TO THE FILE EXPLOITER!!!!

Answer:

I think B

Explanation:

Red litmus paper is used more to find bases, blue litmus paper changes better when coming to find acids. So I believe it is B

The best approach best investigate whether substances are basic is to test each substance with red litmus paper.

Basic compounds are those which have a pH value in between the range of 7.1 to 14.

Red litmus paper changes its color from red to blue when comes in the contact of basic compounds. So to check the basicity of household substance we should use red litmus paper.

Hence test each substance with red litmus paper.

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thats kinda hot not even gon'

The most electrically conductive solution among the three given solutions is hydrobromic acid, which ionizes completely in the solution and produces a high amount of ions. Therefore, option (b) hydrobromic acid is the correct answer to this question.

Acids produce ions in solution, which leads to the solution being more conductive. The more ions an acid produces, the higher the electrical conductivity of the solution. Hence, to determine the electrical conductivity of the acid, we need to know the number of ions generated by the acid in the solution. Here, three acids: hydrobromic acid, acetic acid, and chloric acid have been taken at the same concentration. So, let's check the number of ions produced by each of the acids: Hydrobromic acid: It is a strong acid that dissociates completely in the aqueous solution.

The dissociation reaction is: HBr → H+ + Br−Since it ionizes completely in the solution, the electrical conductivity of the solution would be high. Acetic acid: It is a weak acid that dissociates partially in the aqueous solution. The dissociation reaction is: CH3COOH ↔ H+ + CH3COO−Since it does not ionize completely in the solution, the electrical conductivity of the solution would be low compared to hydrobromic acid. Chloric acid: It is a strong acid that dissociates completely in the aqueous solution. The dissociation reaction is: HClO3 → H+ + ClO3−Since it ionizes completely in the solution, the electrical conductivity of the solution would be high. So, the most electrically conductive solution among the three given solutions is hydrobromic acid, which ionizes completely in the solution and produces a high amount of ions. Therefore, option (b) hydrobromic acid is the correct answer to this question.

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Answer:

Newtons second law says that acceleration (gaining speed) happens when a force acts on a mass (object)

Explanation:

For example riding your bicycle, your bicycle is the mass and your leg muscles pushing on the pedals of your bicycle is the force.

In the given reaction, 4 moles of KO2 react to produce 3 moles of O2. Therefore, if 10.5 moles of carbon dioxide (CO2) are used at STP, an equal amount of oxygen (O2) is produced. Hence, 10.5 moles of O2 will be generated.

According to the balanced chemical equation, the reaction 4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g) shows that for every 4 moles of KO2, 3 moles of O2 are produced. Therefore, the molar ratio between KO2 and O2 is 4:3.

Since 10.5 moles of CO2 are used, we can use the stoichiometry of the reaction to determine the amount of O2 produced. As the reaction equation does not involve CO2, the amount of CO2 used does not directly affect the production of O2. Thus, 10.5 moles of CO2 will yield an equal amount of O2, resulting in 10.5 moles of O2 being produced at STP.

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The different parts of the chemical equation below are identified as follows:

1. A: Solid

2. B: Liquid

3. C: Gas

4. D: Aqueous

5. (s): Solid

6. (l): Liquid

7. (g): Gas

A chemical equation is a symbolic representation of a chemical reaction where reactants are written on the left side and products on the right side. These reactants and products are identified with their respective states such as solid, liquid, gas, and aqueous.

The symbols for the physical states of the reactants and products are included in parentheses, alongside their respective chemical formulas.

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Answer:

Becxause there is no whee5r else for the flame to go out fromk

Explanation:

NaCl

Na = 1

Cl=1

Answer:

19. because fire needs oxygen or air to keep the fire burning.

20.A molecule of sodium chloride, NaCl, consists of one atom each of sodium and chlorine. Hence, each molecule of NaCl has 2 atoms total. Similarly, we can calculate the total number of atoms in 1 mole of sodium chloride. Since 1 molecule contains 1 atom each of Na and Cl, 1 mole of NaCl will contain 1 mole each of Na and Cl.

Explanation:

Hope this helps!

Answer:

same I was looking for a question like this I was asking question and someone kept commenting this bitly thing i dont kno I though it was just weird and frustrating lol

Unicellular organisms such as amoeba and leishmania undergo binary fission.

Binary fission can be described as a form of asexual reproduction where an organism divides into two, each part having one copy of genetic material.

Binary fission can be described as a kind of asexual reproduction observed in prokaryotes and a few single-celled eukaryotes. In asexual reproduction, there is a separation of the parent cell into 2 new daughter cells. This process occurs with the division and duplication into two parts.

Binary fission is a primary of reproduction in prokaryotic organisms. Binary Fission happens without spindle apparatus formation in the cell.  The single DNA molecule starts replication and attaches each copy to parts of the cell membrane.

Prokaryotes and eukaryotes such as euglena reproduce via binary fission. This process is rapid, and its speed varies in species. The time needed by bacteria to double the cells it has is called known as doubling time.

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The mass of the HCl that is required for the reaction is 2.92 g

The mass of the hydrogen gas produced is 0.08 g

Number of moles of the iron = 2.3 g/56 g/mol

= 0.04 moles

We have that 1 mole of the iron reacts with 2 moles of HCl

0.04 moles of iron would react with 0.04 * 2/1

= 0.08 moles

Mass of the HCl required = 0.08 moles * 36.5 g/mol

= 2.92 g

We have that the mole ratio of the iron and the hydrogen is 1:1 kit then follows that mass of the hydrogen produced would now be;

0.04 moles * 2 g/mol

= 0.08 g

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YESS It is true!!. In general, atoms are most stable, least reactive, when their outermost electron shell is full. Most of the elements important in biology need eight electrons in their outermost shell in order to be stable, and this rule of thumb is known as the octet rule.

Answer:

It's 0.00015

Explanation:

got it right on edge 2021

Answer:

for the first one we need the previous question to be able to answer it. for the second one density and temperature

Explanation:

Answer:

4- radioactive isotopes

Explanation:

I don't remember exactly but this question was on the regents

Uranium-238  is a non- fissile element. It is a radioactive isotope that can be best used to determine geological formations. Thus, option 4 is correct.

Radioactive isotopes are elements that have an unstable atomic nucleus and can undergo radioactive decay to produce new particles and energy. They have the same atomic number as that of their parent species.

Isobars are a substance that has atomic mass and do not include uranium. Nuclear fusion and fission are the processes of nuclear energy that combine or splits the unstable nucleus to form a new particle.

Therefore, option 4. Uranium-238 is a radioactive isotope.

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The major E2 reaction product formed when (1R,2S,4S)-2-chloro-4-methyl-1-(propan-2-yl)cyclohexane reacts with hydroxide ion in DMSO is (1S,2R,4S)-2-methyl-1-(propan-2-yl)cyclohexene.

In an E2 (elimination) reaction, a hydrogen atom and a leaving group are removed from adjacent carbon atoms, resulting in the formation of a double bond. In this case, the leaving group is the chloride ion (Cl-) and the hydrogen atom is removed from the carbon adjacent to the chlorine atom.

The given starting compound, (1R,2S,4S)-2-chloro-4-methyl-1-(propan-2-yl)cyclohexane, has a chiral center at carbon 2, which is marked as (S). When the hydroxide ion (OH-) attacks the carbon adjacent to the chlorine atom, the chlorine leaves and a hydrogen from the adjacent carbon is also removed. The resulting product is (1S,2R,4S)-2-methyl-1-(propan-2-yl)cyclohexene.

It's important to note that the stereochemistry of the starting compound and the resulting product is represented by the (R) and (S) designations, indicating the arrangement of the substituents around the chiral carbon atoms.

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The balanced equation is 6 Pb(s) + 6 Cr2O72-(aq) + 42 H+(aq) → 12 Cr3+(aq) + 42 H2O(l) + 6 Pb2+(aq). 6 moles of Pb(s) are oxidized for every 6 moles of Cr2O72- consumed. If 5 moles of Cr2O72- are used, it will oxidize 5 moles of Pb(s).

First, we need to balance the two half-reactions. Let's start with the reduction half-reaction of Cr2O72-: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l).  To balance the oxygen atoms, we add 7 H2O to the left side: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l). Next, we balance the hydrogen atoms by adding 14 H+ to the right side:

Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) + 14 H+(aq)

Now, let's balance the oxidation half-reaction of Pb: Pb(s) → Pb2+(aq) + 2 e-. The equation is already balanced. To combine the two half-reactions, we multiply the reduction half-reaction by 6 and the oxidation half-reaction by 6 to ensure that the electrons cancel out: 6 Cr2O72-(aq) + 42 H+(aq) + 36 e- → 12 Cr3+(aq) + 42 H2O(l) + 84 H+(aq). 6 Pb(s) → 6 Pb2+(aq) + 12 e-. Now, we can write the balanced equation: 6 Pb(s) + 6 Cr2O72-(aq) + 42 H+(aq) → 12 Cr3+(aq) + 42 H2O(l) + 6 Pb2+(aq).

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Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

[tex]Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J[/tex]

Thus, the total energy turns out to be:

[tex]Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J[/tex]

Best regards!

By dividing the concentration of the reference solution by 100, you will obtain the concentration of the desired solution, which is 100 times less concentrated than the reference solution.

To find the concentration of a solution that is 100 times less than another solution, you can follow these steps:

1. Determine the concentration of the reference solution. Let's denote it as [A]₀.

2. Calculate the concentration of the desired solution. Let's denote it as [A].

3. Since the desired solution is 100 times less concentrated than the reference solution, you can divide the concentration of the reference solution by 100 to obtain the concentration of the desired solution:

  [A] = [A]₀ / 100

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