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3 years ago

If 25.0 mL of 0.100 M Ca(OH)2 is titrated with 0.200 M HNO3, what volume of nitric acid is required to neutralize the base?

Chemistry

1 answer:

Troyanec [42]3 years ago

7 0

Answer:

25 mL

Explanation:

The reaction that takes place is:

Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O

First we calculate how many Ca(OH)₂ moles were spent in the titration:

25.0 mL * 0.100 M = 2.5 mmol Ca(OH)₂

Then we convert Ca(OH)₂ moles into HNO₃ moles, using the stoichiometric ratio:

2.5 mmol Ca(OH)₂ * $\frac{2mmolHNO_3}{1mmolCa(OH)_2}$ = 5.0 mmol HNO₃

Finally we calculate the volume of required nitric acid solution, using the concentration:

5.0 mmol ÷ 0.200 mmol/mL = 25 mL

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highly endothermic

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2 years ago

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Answer:

In the explanation

Explanation:

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3 years ago

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC

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Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$ *100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$ *100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$ *100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$ *100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$ *100%

%IC = 55.5%

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Answer:

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