* I need a answer and equations used in every temrs. ( hand writing preffered )

Question

* I need a answer and equations used in every temrs. ( hand writing preffered )

1) A ballon is filled wit hgas and is heated from 40.0 C to 80.0 C at constant pressure. if the original volume of the ballon was 2.9 L what is the voulme of the baloon after heating?

2) A sample of hydrogen gas occupies 1.75 L at 1.2 atm and 32 C wha pressure in ( atm ) with the sample exist at 3.50L and 59 C?

3) suppose that in a known volume of He, these are 6.022*10^23 atoms at 1.00 atm and 0 C. how many moles and how many grams of F2 are these under the same condition? _____moles and ____grams

4) what is the pressure (in atm) exerted by 90.0g sample of cl2 that occupies 3.20 L at 32.5 C?

5) a gas is inside cylinder with a postion.

a) what is the increase in pressure if the Kelvin Temp increased by 300% ( constant moles , and volumes)

b) what is the chnage in the pressure if the volume decreased by 60%?

6) if 2.00ml o gas at STp has a mass of 1.60g. what is its molar mass?

7) a bomb calorimeter have heat capacity of 795 J/C and contains 864 grams of water. if the combuistion of 0.650 mole of hydrocarbon increases the temperature of its calorimeter and the water from 25.3C to 55.7C determine the enthalpy eper mole of the hydrocarbon. Delta H ______ KJ/mole

Explanation / Answer

1) since pressure is kept constant, therefore, Volume is directly proportional to temperature (Boyle's Law)

thus, V1/T1 = V2/T2

or, 2.9/(273+40) = V2/(273+80)

or, V2 = 3.271 litres

2) since number of moles of the gas remains constant, therefore

applying {(P1)*(V1)}/(T1) = {(P2)*(V2)}/(T2)

or, {1.2*1.75}/(273+32) = {(P2)*3.5}/(273+59)

or, P2 = 0.653 atm

3) 6.022*10^23 atoms of any species corresponds to 1 moe of that species

thus, moles of F2 = 1

molar mass of F2 = 38 g/mole

thus, mass of F2 = 38 g

4) molar mass of Cl2 = 71 g/mole

therefore, moles of Cl2 in 90 g of it = 90/71 = 1.268

applying P*V = n*R*T

or, P = n*R*T/V = 1.268*0.0821*(273+32.5)/3.2 = 9.935 atm

5) a) under constant moles and constant volumes

pressure directly proportional to temperature

thus, increase in pressure = 300%

b) under constant moles and constant temperature,

pressure is inversely proportional to volume

therefore, pressure will increase by 60%

6) at STP , 22.4 litres of any gas corresponds to 1 mole of it

therefore, 2 ml corresponds to 2/22400 = 9.93*10^-5 moles of the gas

now, molar mass = mass of 1 mole

thus, if  9.93*10^-5 moles of the gas weighs 1.6 g

therefore, 1 mole of the gas weighs 1.6/(9.93*10^-5) = 17920 g = molar mass of the gas

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