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Leona [35]
3 years ago
9

Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.11 M in sodium lactate

Chemistry
1 answer:
Gelneren [198K]3 years ago
5 0
Acidity of lactic acid = pKa = 3.85
Ca = 0.13M
Cs = 0.11M
<span>Henderson-Hasselbalch equation:</span>
pH = pKa + log(Cs/Ca)
pH = 3.85 + log(0.11M/0.13M) = 3.85 + log0,846 = 3.85 - 0.073 = <u>3.777</u>
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Ozone, O3(g), is a form of elemental oxygen produced during electrical discharge. Is ΔH∘f for O3(g) necessarily zero? Yes or no
Ivan
The answer for your question is <span>No. This is because in given conditions, it is not the most stable form of oxygen's element. It will not equate into zero because there will be charge remained after balancing the equation. 
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6 0
4 years ago
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Which can be excluded due to the absence of fossil records
xeze [42]

Answer:

a. body of jellyfish

Explanation:

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8 0
3 years ago
A mixture of helium and neon gases has a density of 0.6048 g/L at 48.7°C and 792 torr. What is the mole fraction of neon in this
weeeeeb [17]
The volume of one mole of gas is 22.4 liters at 0 °C and 760 Torr. The volume of the given condition can be calculated using:
(PV)/T = constant
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4)/273 = (792 * V₂) / 321.7
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The Mr of Ne = 20
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1 - x is the fraction of He
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0.6048 = (20x + 4(1 - x))/25.3
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5 0
3 years ago
Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titra
Setler79 [48]

Complete Question:

Consider these three titrations: (i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH (ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH (iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH. Which statement is most likely to be true?

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

(b) All three titrations have the same initial pH.

(c) All three titrations have the same pH at their first equivalence point.

Answer:

(a) All three titrations require the same volume of NaOH to reach their first equivalence point.

Explanation:

(i) the titration of 25.0 mL of a 0.100 M monoprotic weak acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

ii) the titration of 25.0 mL of a 0.100 M diprotic weak acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

                     

(iii) the titration of 25.0 mL of a 0.100 M strong acid with 0.100 M NaOH

number of moles of acid = \frac{25}{1000} dm^{3}  * 0.1 M = 0.0025 moles

Therefore, all the acids require the same number of moles of NaOH to reach their first equivalence points

Note that the concentration of the base NaOH are also the same, therefore the volume of NaOH required to reach equivalence point would also be the same for all the three titrations.

All three titrations don't have the same initial and equivalence point pH because they all have different acidic properties.

6 0
4 years ago
Read 2 more answers
A certain substance X has a normal boiling point of 134.5°C and a molal boiling point elevation constant =Kb1.36·°C·kgmol−1 . Ca
erastova [34]

Answer:

136.5°C

Explanation:

Given data

  • Boiling point of X: 134.5°C
  • Molal boiling point elevation constant (Kb): 1.36·°C·kg.mol⁻¹

We can calculate the elevation in the boiling point (ΔT) using the following expression.

ΔT = Kb × m

where,

m is the molality

The molar mass of urea is 60.06 g/mol. The moles of urea corresponding to 76 g is:

76 g × (1 mol/60.06g) = 1.3 mol

The mass of the solvent (X) is 850 g = 0.850 kg.

The molality is:

m = 1.3 mol / 0.850 kg = 1.5 mol/kg

Then,

ΔT = Kb × m

ΔT = 1.36·°C·kg.mol⁻¹ × 1.5 mol/kg

ΔT = 2.0 °C

The boiling point of the solution is

134.5°C + 2.0°C = 136.5°C

3 0
4 years ago
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