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Umnica [9.8K]
4 years ago
13

In a coffee-cup calorimeter, 50.0ml of 0.100M AgNO3 and50.0ml of 0.100M HCl are mixed to yield the following reaction:

Chemistry
1 answer:
Marina CMI [18]4 years ago
5 0

Answer:

-66.88KJ/mol

Explanation:

It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:

q = -C×m×ΔT

<em>Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)</em>

Replacing:

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

If 0.0050 moles produce -334.4J. Heat of reaction is:

-334.4J / 0.0050moles = -66880J/mol = <em>-66.88KJ/mol</em>

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A system is composed of 7.14 grams of Ne gas at 298 K and 1 atm. When 2025 joules of heat are added to the system at constant pr
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Answer:

(a) 1 atm, 8.72 L and 298 K respectively.

(b) 1 atm, 16.72 L and 573.35 K respectively.

(c) \Delta U=1215J

Explanation:

Hello,

(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\  \\V=8.72L

Thus, initial conditions of pressure volume and temperature are 1 atm, 8.72 L and 298 K respectively.

(b) Since the process was carried out at constant pressure, the work is defined as:

W=P(V_2-V_1)

Thus, the final volume is:

V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L

And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:

Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K

Therefore, final volume is 16.72 L, final pressure is also 1 atm and final temperature is 573.35 K for this expansion process.

(c) Finally, the change in the internal energy is computed via the first law of thermodynamics:

Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J

Best regards.

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