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marusya05 [52]
3 years ago
10

What is the temperature of 0.750 mol of a gas stored in a 6,850 mL cylinder at 2.21 atm? Use PV=nRT and R=0.0821 (L•atm)/(mol•K)

Chemistry
2 answers:
nadya68 [22]3 years ago
3 0
Using the formula PV=nRT
we know:
n= 0.750mol
R= 0.0821 (L•atm)/(mol•K)
V= 6850 ml = 6.85 L
P= 2.21 atm
and we need to find T
all you have to do is divide PV by nR and you will find T, like this:
t =  \frac{p \times v}{n \times r} \\ t =  \frac{2.21atm \times 6.85l}{0.750mol \times 0.0821 \frac{atm \times l}{mol \times k} }  \\ t = 246k
The temperature of the gas is D)246 K
SIZIF [17.4K]3 years ago
3 0

Answer:

d.) 246 K

Explanation:

this is the answer on edge, i just took quiz

hope this helps!!!

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How many moles of methanol must be added to 4.50 kg of water to lower its freezing point to -11.0 ∘c? for each mole of solute, t
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<u>Given:</u>

Mass of solvent water = 4.50 kg

Freezing point of the solution = -11 C

Freezing point depression constant = 1.86 C/m

<u>To determine:</u>

Moles of methanol to be added

<u>Explanation:</u>

The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:

ΔTf = kf*m

where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution

ΔTf = 0 C - (-11.0 C) = 11.0 C

m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg

11.0 = 1.86 * moles of methanol/4.50

moles of methanol = 26.613 moles

Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.



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