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3 years ago

ANSWER ASAP I WILL GIVE BRIANIEST

Chemistry

2 answers:

tiny-mole [99]3 years ago

7 0

Answer : The charge on magnesium is (+2) and the the charge on bromide is (-1).

Explanation :

Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.

Ionic compound are usually formed when a metal reacts with a non-metal.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Magnesium bromide is an ionic compound because magnesium element is a metal and bromide element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The charge on magnesium is (+2) and the the charge on bromide is (-1). Both combine with the an ionic bond by the criss-cross method.

Thus, the formula of the compound magnesium bromide will be $MgBr_2$

fredd [130]3 years ago

4 0

Answer:

I think two bromine anions have to combine with one magnesium cation to form magnesium bromide (MgBr2)

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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]

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

Explanation:

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:

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

Initial: 0.20

At eqllm: 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:

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

Initial: 0.083

At eqllm: 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 $8.3064\times 10^{-2}M$

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kolbaska11 [484]

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<h3>Further explanation</h3>

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