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Dmitry_Shevchenko [17]
4 years ago
15

What is revealed in the reaction ca2+(aq) + 2oh-(aq) ca(oh)2(s), deltah + 16.71?

Chemistry
2 answers:
ch4aika [34]4 years ago
8 0
The given reaction has a positive value of enthalpy change which means that it is an endothermic reaction. The heat energy flows into the reaction as it proceeds. So, to allow this reaction to occur heat should be supplied to the substances.
Veronika [31]4 years ago
8 0

The reaction is endothermic - apex


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When 33 g of CaO and 10 g of H2O react, how many grams of calcium hydroxide would you expect to be produced* Explain your answer
MrMuchimi

Answer:

41.44 g

Explanation:

First of all, we must put down the equation of the reaction;

CaO + H2O ----->Ca(OH)2

Number of moles of CaO = 33g/56 g/mol = 0.59 moles

Number of moles of H20 = 10g/18 g/mol = 0.56 moles

Since the reaction is in 1:1 mole ration, H2O is the limiting reactant

Hence;

mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g

7 0
3 years ago
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What is the reduction half-reaction for the following unbalanced redox equation?
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First find the oxidation states of the various atoms: 
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<span>Which of the following is an oxidation half-reaction? </span>
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What is percentage composition?
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7 0
3 years ago
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
bija089 [108]

<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

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b. o2
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4 0
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