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Lelu [443]
3 years ago
10

A sample of N2(g) was collected over water at 25°C and 730 torr in a container with a volume of 340 mL. The vapor pressure of wa

ter at 25°C is 23.76 torr. What mass of N2 was collected?
A. 0.58 g
B. 0.72 g
C. 0.91 g
D. 0.36 g
E. 0.080 g
Chemistry
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

D. 0.36 g

Explanation:

When a gas is collected over water, the total pressure is the sum of the pressure of the gas and the pressure of the water vapor.

Ptotal = Pwater + PN₂

PN₂ = Ptotal - Pwater = 730 torr - 23.76 torr = 706 torr

We can find the mass of N₂ using the ideal gas equation.

P.V=n.R.T=\frac{m}{M} .R.T\\m=\frac{P.V.M}{R.T} =\frac{730torr.0.340L.28g/mol}{(0.08206atm.L/mol.K).298K} .\frac{1atm}{760torr} =0.36g

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What mass (in grams) of Mg(NO3)2 is present in 151 mL of a 0.350 M solution of Mg(NO3)2?
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Answer is: A) 7.84 g.

V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.

V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.

c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.

n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).

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M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.

m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).

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Answer:

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To find ΔH we need to obtain moles of reaction and heat released in the reaction because ΔH is defined as heat released per mole of reaction.

<em>Moles of reaction:</em>

Moles of Ag⁺ and Cl⁻ added are:

Ag⁺: 0.100L * (0.100mol / L) = 0.01moles

Cl⁻: 0.100L * (0.200mol / L) 0 0.02 moles

That means limiting reactant is Ag⁺ and moles of reaction are 0.01 moles

<em>Heat released:</em>

To find heat released we must use coffe cup calorimeter equation:

Q = C*m*ΔT

<em>Where C is specific heat of solution (4.18J/g°C), m is the mass of solution (200g because there are 100 + 100mL = 200mL and density of solution is 1g/mL) and ΔT is change in temperature (25.30°C - 24.60°C = 0.70°C).</em>

Replacing:

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ΔH = 596.05J / 0.01 moles =

ΔH = 59605J / mol =

<h3>ΔH = -59.6kJ/mol</h3>

<em>As heat is released, ΔH < 0.</em>

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