49 and 50 please. Calculate the volume occupied by 35.2 g of methane gas (CH_4)

Question

49 and 50 please. Calculate the volume occupied by 35.2 g of methane gas (CH_4) at 25.0 degree C and 1.00 atm. R = 0.0821 L atm/K mol. Which one of the following substances, will NOT display H-bonding? Consider 1.00 L of air in a patient's lungs at 37.0 degree C and 1.00 atm pressure. What volume would this air occupy if it were at 25 degree C under a pressure of 5.00 Times 10^-2 atm (a typical pressure in a compressed air cylinder)? Which substance is expected to have the highest melting point? Which of the following particles or rays requires barriers of lead and/or concrete for adequate protection? What volume of CO_2 gas at 645 torr and 800 K could be produced by the reaction of 45.0 g of CaCO_3 according to the equation?

Explanation / Answer

Solution :-

Q49).

The penetration power of the gamma particle is very high therefore it needs the partition of the concreate or lead

so answer is option C

Q50).

CaCO3 ------ > CaO + CO2

Lets first calculate the moles of the CaCO3

Moles = mass / molar mass

  Moles of CaCO3 = 45.0 g / 100.09 g per mol = 0.4496 mol CaCO3

1mol CaCO3 gives 1 mol CO2 gas

Therefore moles of CO2 that can be formed = 0.4496 mol CO2

Now lets calculate the volume of the CO2

PV= nRT

V= nRT/P

   = 0.4496 mol * 0.08206 L atm per mol K * 800 K / (645 torr * 1 atm /760 torr)

   = 34.8 L

So the CO2 that can be formed is 34.8 L that is option D

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