A sample of gas occupies 30.0 liters at STP. What will be its temperature in deg

Question

A sample of gas occupies 30.0 liters at STP. What will be its temperature in degree C if the volume is reduced to 22.5 liters under 3.50 atmospheres pressure? 212 degree C 284 degree C 386 degree C 444 degree C 516 degree C What is the pressure of 64.0 g of oxygen gas in a 1.50-L container at -37 degree C? 4.12 atm 25.8 atm 51.6 alm 19.6 atm 8.2 atm What is the volume occupied by 10.0 grams of NH_3 at 500 degree C and 1140 torr? 14.9 L 16.1 L 24.9 L 32.7 L 46.2 L What volume of NO (STP) can be produced by dissolving 62.1 grams of lead in nitric acid? 3Pb + 8HNO_3 rightarrow 3Pb(NO_3)_2 - 2NO + 4H_2O 1.12 L 1.68 L 2.24 L 3.36 L 4.48 L

Explanation / Answer

Q1a) Answer is D

Calculations,

Use P1V1/T1 = P2V2/T2
1atm*30.0L/273 K = 3.50* 22.5 L/ T2
T2 = (273 x 3.50 x 22.5)/30

= 21498.75/30

= 716.625 K
=716.625 - 273

= 443.625

= 444

Q1b) Answer is B.

The ideal gas equation is PV = nRT ;        P = nRT/V

V = 1.5 L ; R =0.08314 L.bar/mol.K.
No. of moles, n = 64/32 = 2
T = 273.15 +(-37) = 236.15 degree K

Therefore, P = 2 x 0.08314 x236.15 /1.5
                     = 26.17 Pa

= 28.5 atm

Q1c) Answer is C.

The ideal gas equation is PV = nRT

No. of moles n=10/17 =0.588

T=273.15+500=773.15

P=1140 torr = 151987.5

V=nRT/P = 0.588 * 0.08314*773.5 / 151987.5 = 2.48 *10-4

V= 24.9L

Q1d) Answer is E

3Pb + 8HNO3 3Pb(NO3)2 + 2NO + 4H2O

assign equation weights/volumes
3x207g 2x22.4L

L of NO = 62.1 g Pb x 44.8 L NO / 621 g Pb = 4.48 L

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