Question

31. Calculate the number of moles and the mass of gas in 3.40 L of He at STP. A. 0.152 mol 0.608 g B. 76.2 mol. 305 g C. 0.152 mol, 0.0379 g D. 76.2 mol. 19.0 g E. 1.00 mol, 4.00 g 32. Calculate the number of moles in 95.0 g of O., and the volume that it would occupy at 80.0°C and 3.50 atm A 11.9 mol, 266 L B. 5.94 mol, 11.1 L C. 5.94 mol, 49.2 L D. 2.97 mol, 24.6 L E. 2.97 mol, 5.57L 33. Ifa 4,00 L container is filled with N, to a pressure of 895 torrat 53.0"C calculate the mass of the nitrogen in the container A. 0.176 g B. 4.93 g C. 134 g D. 5.68 8 E. 0 202 g 34. What volume of Hz would be collected at 21.5°C and a pressure of 695 torr if 5.25 g of zinc react according to the equation: Zn(s) + 2HCl(aq) → ZnCl(aq) + H2(g) A. 1.06 L B 4.25 L C. 1.80 L D 2.12 L

answer all please

Answer 1

31- We know 1 mole of gas has a volume of 22.4l at stop condition.

So 3.40L will be =.3.40/22.4 moles

= 0.152 moles

Now mass of this moles of helium is= 0.152× 4 grams

= 0.608 grams.

So the answer is A.

32- mass of 1 mole of O2 gas = 32 grams

So 95 grams= 95/32= 2.97 moles

Using the ideal gas law to find the volume.

PV= nRT

V= nRT/P

= 2.97×0.0820×353/3.50

= 24.6 liters

So the answer is.

33- we know 1torr= 0.00131578947atm

So 895 torr= 1.1776 atm

Now using ideal gas law= PV= nRT

n= PV/RT

= 1.1776× 4/326×0.0820

= 0.176 grams

So the answer is A.

34-The pressure = 695torr= 0.9145 atm

Moles of zinc used= 5.25/65.38

= 0.080 moles

Now using the ideal gas law= V= nRT/P

= 0.080× 0.0820×294.5/ 0.9145

= 2.12 liters

So the answer is D