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VARVARA [1.3K]
2 years ago
7

Two solids of identical mass, A and B, are analyzed using identical calorimeters. Each calorimeter contains the same amount of w

ater and is at room temperature. When the solids are heated to the same initial temperature and placed in their calorimeters, the final temperature of solid A's calorimeter is higher than that of B. What can we infer from this
Chemistry
1 answer:
DIA [1.3K]2 years ago
5 0

Answer:

Specific heat of solid A is greater than specific heat of solid B.

Explanation:

In the calorimeter, as the temperature is increasing, the vibrational kinetic energy will increase and this means that additional amount of energy will be needed to increase the temperature by the same value. Therefore, we can conclude that specific heat increases as temperature increases.

Now, we are told that the final temperature of solid A's calorimeter is higher than that of B.

This means from our definition earlier, Solid A will have a higher specific heat that solid B.

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The answer to your question is,

B. Government.

-Mabel <3

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3 years ago
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Soluble ionic compounds that dissociate naerly completely when dissolved are classified as _______.
Anettt [7]

The answer is D. Strong electrolytes

Strong electrolyte is a solute or solution that completely or almost  completely dissociates when in solution. These are good conductors of electricity only  when in molten/aqueous solution.

Strong electrolyte(aq) → Cation+(aq) + Anion−(aq)


6 0
2 years ago
Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)
scoray [572]

Answer:

As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)

Explanation:

Oxidation: As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)

  • Balance As: As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)
  • Balance H and O in acidic medium: As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)
  • Balnce charge: As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)......(1)

Reduction: NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)

  • Balance N: 2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)
  • Balance H and O in acidic medium: 2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)
  • Balance charge: 2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)......(2)

Equation (1)+Equation (2) gives-

As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)

3 0
3 years ago
Read 2 more answers
If you remove the cap of a bottle of a soft drink carbon dioxide escapes the soft drink but the soft drink is still a solution.
Effectus [21]

Answer:

Removing the cap from a soft drink bottle releases pressure and causes the excess carbon dioxide molecules to come out as bubbles. however the drink is still supersaturated , and will release that carbon dioxide till it goes flat

6 0
2 years ago
In a zero order reaction, it takes 342 seconds for 75% of a hypothetical reactant to decompose. Determine the half-life t_{1/2}
stiv31 [10]

Answer:

228 s

Explanation:

In a zero order reaction, the formula for the half life is given as;

t1/2 = [A]o / 2k

To obtain the rate constant k, we have to use;

[A] = [A]o - kt

kt = [A]o - [A]

From the question;

it takes 342 seconds for 75% of a hypothetical reactant to decompose.

We have;

t = 324

[A] = 25

[A]o = 100

Upon solving for k we have;

kt = [A]o - [A]

k = ( [A]o - [A] ) / t

k = (100 - 25 ) / 342

k = 75 / 342 = 0.2193

Solving for t1/2;

t1/2 = [A]o / 2k

t1/2 = 100 / 2(0.2193)

t1/2 = 100 / 0.4386 = 228 s

8 0
3 years ago
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