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3 years ago

Two boxes are filled with air. The molecules that make up the air in Box A

Chemistry

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

box a is warmer than box b, so the molecules are moving around a lot more

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Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5

77julia77 [94]

<u>Answer:</u> The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

<u>Explanation:</u>

We are given:

Concentration of buffer = 0.460 M

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

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$ .....(1)

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COOH]=0.460M$

$[C_6H_5COO^-]=0.460M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2$

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

$pK_a$ = 4.2

pH = 3.2

Putting values in equation 1, we get:

$3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

$0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Let the moles of acid added to carry out the change is 'x' moles

Calculating the moles of acid added:

$\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434$

Calculating the volume of acid added by using equation 2, we get:

Moles of acid added = 0.0434 moles

Molarity of solution = 0.390 M

Putting values in equation 2, we get:

$0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL$

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

7 0

3 years ago

Express the frequency in inverse seconds of n=4 to n=3

HACTEHA [7]

Answer:

1.60×10¹⁴ s⁻¹

Explanation:

When an electron jumps from one energy level to a lower energy level some energy is released in the form of a photon.

4 0

3 years ago

The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reac

Mariulka [41]

Explanation:

Al=10,o2=5

(Al10)2(o5)3

The answer is oxygen.

6 0

3 years ago

Edmentum

goldenfox [79]

Answer:

They all contain switches

6 0

2 years ago

What happens to acetone molecules when you add heat to a beaker of liquid acetone? They gain a higher average kinetic energy. Th

iVinArrow [24]

When you heat up most substances it gives them more Kinetic energy and the substance becomes less arranged in an ordered state, further apart and move faster. therefore the answer is the first: They gain a higher average kinetic energy

Hope that helps :)

7 0

3 years ago