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Lina20 [59]
3 years ago
7

HELP PLSSSSS!!

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

68133080.02 g

Explanation:

I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.

Now, if 1 mole of a gas occupies 22.4 L

x moles of air occupies 52,681,428.8 Liters

x = 1 * 52,681,428.8 /22.4

x = 2351849.5 moles of air

Now, number of moles = mass/ molar mass

but molar mass = 28.97 g/mol

2351849.5 = mass/28.97

mass = 2351849.5 * 28.97

mass = 68133080.02 g

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of
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None of the additions will exceed the capacity of the buffer.

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As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

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pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

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Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

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# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

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