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Anastasy [175]
2 years ago
12

When two gases of a chemical reaction are both at the same temperature and pressure and they have the same molar volume, what is

their stoichiometric ratio in the chemical equation describing the reaction?
Chemistry
2 answers:
Rudik [331]2 years ago
4 0
When two gases of a chemical reaction are at the same temperature, pressure and molar volume, then the stoichiometric ratio of the gases would be 1 is to 1. Molar volume is the volume of the gas per mole of the gas. Having the same conditions for both gases would mean that they are present with the same number of moles.
monitta2 years ago
3 0

Answer:

a. 1 to 1

Explanation:

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
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<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.306M

[C_6H_5NH_2]=0.418M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

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