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o-na [289]
4 years ago
9

How much Ca(NO3)2 should be weighed out to have 0.650 mol?A) 66.4 g.B) 97.68.C) 107 g.D) 133 g.E) 165 g.

Chemistry
1 answer:
Bogdan [553]4 years ago
6 0

Answer:

C) 107 g

Explanation:

Step 1: Calculate the molar mass of Ca(NO₃)₂

We can calculate the molar mass of Ca(NO₃)₂ by adding the masses of its elements.

M = 1 × Ca + 2 × N + 2 × 3 × O

M = 1 × 40.08 g/mol + 2 × 14.01 g/mol + 6 × 16.00 g/mol

M = 164.10 g/mol

Step 2: Calculate the mass corresponding to 0.650 moles of Ca(NO₃)₂

We multiply the number of moles by the molar mass.

0.650 mol × 164.10 g/mol = 107 g

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Answer:

The degree of dissociation of acetic acid is 0.08448.

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Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

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Initial concentration of the acetic acid = [HAc] =c = 0.00225

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HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

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