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kotegsom [21]
3 years ago
6

A student did an experiment to determine the specific heat capacity of a metal alloy. The student put a sample of the alloy in b

oiling water for several minutes, then quickly transferred the alloy into a calorimeter containing water originally at 25°C. The temperature of the water was monitored over time. The data are given in the graph above.
(a) What is the value of ΔT that the student should use to calculate the value of q, the heat gained by the water?

(b) In terms of what occurs at the particulate level, explain how the temperature of the water increases after the alloy sample is added.

(c) The student claims that thermal equilibrium is reached at time t. Justify the student’s claim. In your justification, include a description of what occurs at the particulate level when the alloy and the water have reached thermal equilibrium.

Chemistry
1 answer:
guapka [62]3 years ago
7 0

Answer:

a) ∆T=T1-T2

b) At the particle level the temperature changes are the result of the added energy causing the particles of water to move more vigorously. Either the particles of solid vibrate more vigorously about their fixed positions or the particles of liquid and gas move about their container more rapidly.

c) The state in which two substances in physical contact do not share any heat energy. The temperature of two substances in thermal equilibrium is said to be the same. Also see thermodynamics.

Explanation:

hope that helped good luck!

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Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

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