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Darya [45]
3 years ago
11

A common laboratory reaction is the neutralization of an acid with a base. When 50.0 mL of 0.500 M HCl at 25.0°C is added to 50.

0 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter, the temperature of the mixture rises to 28.2°C. What is the heat produced by the reaction.
A) 670 J
B) 1300 J
C) 27 kJ
D) 54 kJ
E) > 100 kJ
Physics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

B

Explanation:

In this calorimetry problem, the heat released by the reaction is equal to the heat absorbed by the solution (assumed to have the same specific heat capacity as water, 4.19 Jg⁻¹°C⁻¹).

The formula Q = mcΔt will be used to calculate the heat energy, where m is the mass, c is the specific heat capacity, and Δt is the change in temperature from final to initial.

The volume of solution is (50.0 + 50.0)mL = 100.0mL = 100.0g, since water has a density of 1.00g/mL.

The heat absorbed by the solution is then calculated.

Q = mcΔt = (100.0 g)(4.19 Jg⁻¹°C⁻¹)(28.2°C - 25.0°C) = 1340 J

The closest answer is B) 1300 J. This answer is obtained by including only two significant figures in the answer.

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Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

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where,

P_0 =Atmospheric pressure = 800 mbar

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\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

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P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

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