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nasty-shy [4]
3 years ago
10

Jill is studying the temperature output of an ammonium chloride hand warmer with a calorimeter. What change should she expect to

see in the temperature of the water over time?
A) The water will continue to heat up as the experiment progresses.

B) The water will remain a relatively constant temperature during the experiment.

C) The water will gradually cool off throughout the experiment like the hand warmer.

D) Initially the water will heat up and then it will slowly cool off as the hand warmer does.
Chemistry
1 answer:
ANEK [815]3 years ago
7 0

Answer:

The water will continue to heat up as the experiment progresses.

Explanation:

When a person's fingers are cold or has a muscles ache, one can use chemical hand warmers to heat them up. There are two types of chemical hand warmer products, and they all depend on exothermic chemical reactions to work.

However, ammonium chloride dissolution is endothermic but its crystalization is exothermic. When used as a hand warner with a calorimeter, it is found that the temperature of the water increases steadily due to the crystalization of the ammonium chloride. Hence the answer.

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Molarity of 0.50 mol sugar in 270 mL of solution.
EastWind [94]
<h3>Answer:</h3>

1.85 M

<h3>Explanation:</h3>

<u>We are given;</u>

  • Number of moles as 0.50 mol
  • Volume of the solution is 270 ml

But, 1000 mL = 1 L

  • Thus, volume of the solution is 0.27 L

We are required to calculate the molarity of the solution;

  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing number of moles with the volume.

Molarity = Moles ÷ Volume

In this case;

Molarity = 0.50 moles ÷ 0.27 L

             = 1.85 Mol/L or 1.85 M

Therefore, molarity of the solution is 1.85 M

5 0
3 years ago
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer
lidiya [134]

The question is incomplete, the complete question is

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V

Answer:

only IV and V

Explanation:

If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.

On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.

This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.

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Answer:

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Explanation:

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