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maksim [4K]
3 years ago
12

Calculate the volume of 5.00 mol of helium at 120C and 1520 mm Hg

Chemistry
1 answer:
levacccp [35]3 years ago
4 0

Answer:

80.7 L

Step-by-step explanation:

This looks like a case where we can use the Ideal Gas Law to calculate the volume.

pV = nRT       Divide both sides by p

V = (nRT)/p

=====

Data:

n = 5.00 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = (120 +273.15) K = 393.15K

p = 1520 mmHg × 1 atm/760 mmHg = 2.00 atm

=====

Calculation:

V = (5.00 × 0.082 06 × 393.15)/2.00

V = 161.3/2.00

V = 80.7 L  

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A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0
2 years ago
Significant figures<br> 5.3316 + 6.87 + 37.48
Aleksandr-060686 [28]

Answer:

Explanation:

01: 5.3316+6.87+37.48

02: 12.2016+37.48

03: 49.6816

04: 49.68

Answer: 49.68 (Decimals: 2; Significant Figures: 4)

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5 0
3 years ago
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In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2
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Answer:

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Explanation:

This is the redox reaction:

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As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.

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Mass C₂H₂ needed : 22.165 g

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