575 mmHg na 8.51 A 25.0-g sample of nitrogen, N2, has a vöiui 8.52 A 0.226-g sam

Question

575 mmHg na 8.51 A 25.0-g sample of nitrogen, N2, has a vöiui 8.52 A 0.226-g sample of carbon dioxide, Co2, h 8.53 Mg metal reacts with HCI to produce hydrogen gas. n kelVIIID pressure of 630. mmHg. What is the temperature, H degrees Celsius, of the gas? as a volume of 525 mL and a pressure of 455 mmHg. What is the temperatur in kelvins and degrees Celsius, of the gas? a. What volume, in liters, of hydrogen at 0°C and 1.00 atm b. How many grams of magnesium are needed to prepare Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq) (STP) is released when 8.25 g of Mg reacts? 5.00 L of H2 at 735 mmHg and 18 °C? 8.54 When heated to 350 °C at 0.950 atm, ammonium nitrate dec poses to produce nitrogen, water, and oxygen gases. 2NH,NO3(s)- 2N2(g) + 4H20(g) + O2(g) a. How many liters of water vapor are produced when 25.8 NH NO, decomposes? LEARNING GOAL 8.8 Partia Use Dalton's law of partial pressures to calculate the total pressure of a mixture of gases. Many gas sample mostly oxygen ar cles behave in the result of the colt

Explanation / Answer

8.52

Molar mass of CO2 = 44 g/mol

So, 44 g of CO2 = 1 mol

1 g of CO2 = (1 / 44) mol

0.226 g of CO2 = (0.226 / 44) mol = 0.005 mol

So, n = 0.005 mol

Volume = 525 mL = 0.525 L

Pressure = 455 mmHg = 0.60 atm            ( 1 mmHg = 0.00132 atm)

R = 0.082 L atm mol-1K-1

Now,

PV = nRT

T = PV/nR

   = (0.60 atm) (0.525 L) / { (0.005 mol) (0.082 L atm mol-1K-1) }

   = 768 K

8.53

Mg(s)   +   2HCl(aq)     =     H2(g)   +   MgCl2(aq)

In the reaction equation,

1 mole of Mg produces 1 mole of H2

Now,

Molar mass of Mg = 24 g/mol

So, 24 g of Mg = 1 mol

1 g of Mg = (1/24) mol

8.25 g of Mg = (8.25/24) mol = 0.34375 mol

So, H2(g) will produce = 0.34375 mol

Now,

n = 0.34375 mol

T = 0 oC = 273 K

P = 1 atm

R = 0.082 L atm mol-1K-1

PV = nRT

V = nRT/P

    = (0.34375 mol) (0.082 L atm mol-1K-1) (273 K) / (1 atm)

    = 7.70 L

(b)

V = 5 L

P = 735 mmHg = 0.967105 atm

T = 18 oC = (18 + 273) K = 291 K

R = 0.082 L atm mol-1K-1

Now,

PV = nRT

n = PV/RT

    = { (0.967105 atm) (5 L) } / { (0.082 L atm mol-1K-1) (291 K) }

    = 0.20 mol

So, moles of Mg = 0.20 mol

Molar mass of Mg = 24 g/mol

So, 1 mol of Mg = 24 g

0.20 mol of Mg = 0.2 x 24 g = 4.8 g

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