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AleksAgata [21]
3 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]3 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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How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
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Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

7 0
3 years ago
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