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Vilka [71]
4 years ago
12

Please help me! This is too hard!?

Chemistry
1 answer:
Annette [7]4 years ago
5 0

Answer: 59.24 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.7L

Temperature T1 = 42.7°C

Convert Celsius to Kelvin

(42.7°C + 273 = 315.7K)

Pressure P1 = 684.9 torr

Final Volume V2 = 0.14 L

Final temperature T2 = 803.1°C

Convert Celsius to Kelvin

(803.1°C + 273 = 1076.1K)

Final pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(684.9 torr x 2.7L)/315.7K = (P2 x 0.14L)/1076.1K

1849.23/315.7 = 0.14P2/1076.1

Then, cross multiply

1849.23 x 1076.1 = 315.7 x 0.14P2

1989956.403 = 44.198P2

Divide both sides by 44.198

1989956.403/44.198 = 44.198P2/44.198

45023.67 torr = P2

Now, convert the pressure in torr to atmosphere

If 760 torr = 1 atm

45023.67 torr = Z atm

Then, cross multiply

760 torr x Z = 45023.67 torr x 1 atm

Z = 45023.67 torr / 760 torr

Z = 59.24 atm

Thus, the new pressure of the gas will be 59.24 atm

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245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

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<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

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pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

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