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sergij07 [2.7K]
2 years ago
15

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

sure is 1.350 atm . Part A What is the partial pressure of argon, PAr, in the flask?
Chemistry
1 answer:
tatiyna2 years ago
4 0

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

⇒ P = \frac{m R T}{V}

Put all the values in above formula we get

⇒ P = \frac{0.00115}{0.001} × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

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2 years ago
As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den
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1.01atm is the pressure of the gas

Explanation:

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5 0
1 year ago
The pka of hf is 3.2 determine the pkb of hf?
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Well, first we must remember that

pK_{a}+pK_{b}=14

This is because

K_{a}*K_{b}=10^{-14}

-log(K_{a}*K_{b})=-log(10^{-14})\\-logK_{a}+-logK_{b}=-log(10^{-14})\\pK_{a}+pK_{b}=14

So then

pK_{b}=14-pK_{a}=14-3.2=1.8

7 0
2 years ago
Read 2 more answers
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