1. If you were to make a barometer out of water instead of Hg. how high must the

Question

1. If you were to make a barometer out of water instead of Hg. how high must the column of water be to exert a pressure equal to that of a 760 mm column of Hg? The density of water is 1.0 g/mL and the density of Hg is 13.6 g/mL column must be high (you supply the units). 2. If the gas in a 2.0 L cylinder at 30.0 atm and 25° C is completely transferred into a 15.0 L vessel at 50°C, what will be the pressure of that gas? pressure = atm 3. For the reaction below, how many liters of N2 (g) at 450° C and 6.00 atm are required to produce 1.00 mole of NHs (g)? volume of N2 35°C. What is the partial pressure of each gas? N2(g) + 3H2 (g) 2NH3 (g) 4. A mixture of 0.435 moles of N2, 0.123 moles of O2, and 0.725 mole of Ne is in a 10.0 liter container at 5. A graduated cylinder is filled brim full of water inverted into a beaker of water and used to collect hydrogen produced by a chemical reaction. After the reaction is completed, the volume of gas in the graduated cylinder is 76.0 mL. If the temperature is 30°C and the barometric pressure is 748 torr, how many moles of Hz were collected over the water? You may use your book as necessary moles of H2

Explanation / Answer

2. Using,

moles (n) of gas = PV/RT

with,

P = 30 atm

V = 2 L

R = gas constant

T = 25 oC + 273 = 298 K

So,

moles (n) of gas in cyllinder = 30 x 2/0.08205 x 298 = 2.454 mol

when the gas was transferred to V = 15 L vessel at T = 50 oC + 273 = 323 K

new pressure of gas (P) = nRT/V

                                       = 2.454 x 0.08205 x 323/15 = 4.336 atm

3. moles of NH3 = 1.00 mol

moles of N2 (n) needed = 1.00 mol/2 = 0.5 mol

So,

volume of N2 needed = nRT/P

with,

R = gas constant

T = 450 oC + 273 = 723 K

P = 6 atm

Volume of N2 needed = 0.5 x 0.08205 x 723/6 = 4.94 L

4. Total moles of gases (n) = 0.435 +.123 + 0.725 = 1.283 mol

Total pressure of the system (P) = nRT/V

with,

R = gas constant

T = 35 oC + 273 = 308 K

V = 10 L

So,

P = 1.283 x 0.08205 x 308/10 = 3.24 atm

Partial pressure of a gas = mole fraction of gas x total pressure

Partial pressure of N2 = (0.435/1.283) x 3.24 = 1.10 atm

Partial pressure of O2 = (0.123/1.283) x 3.24 = 0.31 atm

Partial pressure of Ne = (0.725/1.283) x 3.24 = 1.83 atm

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