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shepuryov [24]
3 years ago
6

A gas mixture contains 1.25 g n2 and 0.85 g o2 in a 1.55 l c ontainer at 18 °c. calculate the mole fraction and partial pressure

of each component in the gas mixture.
Chemistry
1 answer:
KatRina [158]3 years ago
8 0

Answer:- mole fraction of nitrogen is 0.626 and mole fraction of oxygen is 0.374.

partial pressure of nitrogen is 0.687 atm and partial pressure of oxygen is 0.410 atm.

Solution:- Moles of nitrogen = 1.25 g x (1mol/28g) = 0.0446 mol

moles of oxygen = 0.85 g x (1mol/32g) = 0.0266 mol

Total moles of gases in the container = 0.0446 + 0.0266 = 0.0712

mole fraction of a gas = moles of gas/total moles of the gases

so, mole fraction of nitrogen = 0.0446/0.0712 = 0.626

mole fraction of oxygen = 0.0266/0.0712 = 0.374

Volume of the container is 1.55 L and the temperature is 18 degree C that is 18 + 273 = 291 K

From ideal gas law equation, PV = nRT

P = nRT/V

Partial pressure of nitrogen = (0.0446 x 0.0821 x 291)/1.55 = 0.687 atm

and the partial pressure of oxygen = (0.0266 x 0.0821 x 291)/1.55 = 0.410 atm

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Answer:

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NHCH3- Withdraws electrons inductively

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Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.

Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

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3 years ago
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4 0
2 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
QveST [7]
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
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Answer:

\huge\boxed{\sf OH-}

Explanation:

<u>According to Arrhenius concept of acid and base:</u>

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

<u>For example:</u>

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\rule[225]{225}{2}

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