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DENIUS [597]
3 years ago
12

For many purposes we can treat nitrogen as an ideal gas at temperatures above its boiling point of - 196.°C.

Chemistry
2 answers:
Slav-nsk [51]3 years ago
4 0

Answer:

Explanation:

P1V1=n1RT1

P2V2=n2RT2

n1=n2  R=R

Divide equals and get

P1V1T2=P2V2T1

V2 =V1 (P1/P2)(T2/T1)

V2 will reduce by 10% due to pressure

V2 will increase by 184K/175K or 5%

therefor overall V will decrease

about 5%

Pavel [41]3 years ago
3 0

Answer: the volume of the sample decreased

Explanation:

T1 = -98°C = - 98 + 273 = 175K

T2 = -89°C = -89 +273 = 184K

P1 = P

P2 = 110%P = 1.1P

V1 = V

V2 =?

P1V1/T1 = P2V2/T2

PxV/175 = 1.1PxV2/184

175x1.1PxV2 = PVx 184

V2 = (PVx 184) /(175x1.1P)

V2 = 0.96V = 96%V

Therefore, the final volume is 96% of the initial volume. This means that the final volume decreased by 96%

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1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
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2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
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0,00675 mol : n(KCl) = 1 : 1.
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n - amount of substance.
M - molar mass.
5 0
3 years ago
When 11.12 g of neon is combined in a 100 L container at 80oC with 9.59 g of argon, what is the mole fraction of argon?
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The correct answer for the question that is being presented above is this one: "<span>0.3."

Here it is how to solve.
M</span><span>olecular mass of Ar = 40
</span><span>Molecular mass of Ne = 20
</span><span>Number of moles of Ar = 9.59/40 = 0.239
</span><span>Number of moles of Ne = 11.12/20= 0.556
</span><span>Mole fraction of argon = 0.239/ ( 0.239 + 0.556) = 0.3</span><span>
</span>
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