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3 years ago

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What is the composition and structure of ionic compounds

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

Sodium and chlorine ions come together to form sodium chloride or NaCl. The sodium atom in this compound loses an electron to become Na+, while the chlorine atom gains an electron to become Cl-. Together, they form a neutral compound because the ions balance each other out.

Explanation:

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C1 Progress quiz: Atomic structure 2 - test

myrzilka [38]

Answer:

0.11 nm

Explanation:

1.1 x 10-10 m

The goal is to convert the atomic radius from meters (m) to nanometres (nm). We do this by multiplying the value by 10^9.

This is given as;

$1.1 * 10^{-10} * 10^{9}\\1.1 * 10^{-10 + 9}\\1.1 * 10^{-1}\\0.11$

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3 years ago

Which of the following is a biotic part of an environment?<br> mud<br> clams<br> sunshine<br> rocks

fredd [130]

Answer:

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Explanation:

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Alinara [238K]

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Where and when are acute-phase proteins produced?

mrs_skeptik [129]

Answer:

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2 years ago

Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

<u>Answer:</u> The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

<u>Explanation:</u>

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

<u>Initial:</u> 0.083

<u>At eqllm:</u> 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago