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AleksAgata [21]
4 years ago
14

Ag2O(s) → 2Ag(s) + ½ O2(g) ΔH° = 31.05 kJ Which statements concerning the reaction above are true? (1) heat is released (2) heat

is absorbed (3) reaction is exothermic (4) reaction is endothermic (5) products have higher enthalpy content than reactants (6) reactants have higher enthalpy content than products A) 1, 3, and 5 B) 1, 3, and 6 C) 2, 4, and 6 D) 2, 4, and 5
Chemistry
1 answer:
Sidana [21]4 years ago
5 0

Answer:

D) 2, 4, and 5

Explanation:

In order to fully comprehend the answer choices we must take a close look at the value of ΔH° = 31.05. The enthalpy change of the reaction is positive. A positive value of enthalpy of reaction implies that heat was absorbed in the course of the reaction.

If heat is absorbed in a reaction, that reaction is endothermic.

Since ∆Hreaction= ∆H products -∆H reactants, a positive value of ∆Hreaction implies that ∆Hproducts >∆Hreactants, hence the answer choice above.

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vazorg [7]

Answer:

1. 46.44 Kj

2. 126.3 J|K

3. 43.34 Kj

Explanation:

change in temperature

25+274 = 298K

100+273 = 373

1.

∆H = nC∆T + n∆H

1x75.5x75+1x40.79

= 5647.5+40.9

we convert 5647.5 to Kj

= 5.6475+40.79

= <em><u>46.44Kj</u></em>

<em><u>2</u></em><em><u>.</u></em>

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<em>∆</em><em>S</em><em> </em><em>=</em><em> </em><u><em>126.3J</em><em>|</em><em>K</em></u>

<em>3</em><em>.</em>

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8 0
3 years ago
How many moles does 13.3 g of Cu contain?
mariarad [96]
<h2>Hello!</h2>

The answer is:

There are 0.209 moles of Cu in 13.g of Cu

Why?

To calculate how many moles does a sample of any element has, we need to use its atomic mass

We are working with Copper (Cu), so we need to find its atomic mass to calculate how many moles does 13.3 g of Cu contains.

So, calculating we have:

Cu=63.54\frac{g}{mol}

We have that there is 1 mol per 63.54 grams of Cu.

Now, converting we have:

13.3g*\frac{1mol}{63.54g}=0.209moles

We have that there are 0.209 moles of Cu in 13.g of Cu

Have a nice day!

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4 years ago
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stealth61 [152]

Answer:

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Explanation:

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E°cell= 2.52 V

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F= Faraday's constant = 96500 C

E°cell = 2.52 V

∆G°=- (4 × 96500 × 2.52)

∆G°= -972720 J

∆G°= -972.72 KJ

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Answer:

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Explanation:

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