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In this experiment, you will fill a gelatin capsule with an alloy
sample. If the mass of the gelatin capsule is 0.123 g, and the mass
of the capsule plus alloy sample is 0.215 g, calculate the
mass, in grams, of the alloy sample. - 0.092 g |

The
temperature of the water is 24.0 |

A. The barometric pressure is 766.1 mmHg. The vapor pressure of water at 24.0 C is 22.4 mmHg. Calculate the pressure, in mmHg, of the dry hydrogen gas.

B. Convert the pressure of the dry H2 to units of atm. 760.0 mm Hg = 1 atm.

C. Calculate moles of H2 produced in the reaction. R = 0.08206 L atm/(mol K)

Answer #1

A .

According to dalton's law total pressure of gas is equal to sum of partial pressure exerted by each individual gas

P_{total} = P_{1} + P_{2}

P_{H2} = P_{total} - P_{water
vapour}

P_{H2} = 766.1 - 22.4 = 743.7 mmHg

**Pressure of dry hydrogen gas = 743.7 mmHg**

B.

760.0 mmHg = 1 atm then 743.7 mmHg = 743.7 X 1 / 760 = 0.97855 atm

**743.7 mmHg = 0.97855 atm**

C.

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT where, P = atm pressure= 0.97855 atm,

V = volume in Liter = 89.5 ml = 0.0895 L

n = number of mole = ?

R = 0.08206L atm mol^{-1} K^{-1}
=Proportionality constant = gas constant,

T = Temperature in K = 297.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.97855X 0.0895) / (0.08206X 297.15) = 0.00359 mole

**mole of H _{2} produced in reaction = 0.00359
mole**

Q1))) A student carried out this experiment and found:
mass gelatin capsule = (1.2010x10^-1) grams
mass alloy sample + gelatin capsule = (2.86x10^-1) grams
mass empty beaker= (1.4083x10^2) grams
mass beaker + displaced water= (2.9086x10^2) grams
temperature= 294.15 K
Barometric pressure= 746 mmHg
vapor pressure of water= 18.65 mmHg
How many liters of hydrogen gas were produced? Enter answer in
scientific notation with three significant figures.
Q2.))) A student performed this lab, but accidentally used the
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