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4 years ago

A chemical bond is formed between magnesium and chlorine. Describe the process and identify the nature of the chemical bond.

Chemistry

1 answer:

IRINA_888 [86]4 years ago

7 0

Answer:

The answer to your question is: b

Explanation:

a. Magnesium shares an electron somewhat unevenly from its 3s orbital with the 3p orbital of chlorine producing a mildly polar covalent bond. This option is wrong because Mg does not share electrons it loses electrons.

b. Magnesium loses and electron from the 3s and gives it up to the 3p of chlorine producing an ionic bond. This option is correct, Mg loses one electron and Cl receives it, the bond formed between Mg and Cl is ionic.

c. Magnesium does not react chemically with chlorine because magnesium gives up electrons, but chlorine only shares electrons. This answer is wrong, Mg and Cl react and produce MgCl₂.

d. Magnesium shares an electron from the 3s orbital with the 3p orbital of chlorine producing a covalent bond. Mg does not share electrons and is not able to produce covalent bonds.

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How many grams of silver nitrate are needed to react with 156.2g of sodium sulfide to produce 595.8g of silver sulfide and 340.0

qaws [65]

Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃

Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate

Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams

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4 years ago

Where does a mid ocean ridge form?

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4 years ago

If a 67.3G rock is dissolved in 2.00L of acid, what is the molar concentration of gold in the acid solution

brilliants [131]

Answer:

[Au] = 0.171 M

Explanation:

For this question, we assume the rock is 100 % gold.

First of all, we determine the moles of gold

67.3 g . 1mol/ 196.97g = 0.342 moles

Molar concentration is defined as the moles of solute, contained in 1L of solution.

Our solution volume is 2L.

M = 0.342 mol / 2L = 0.171

Molar concentration, also called molarity of solution is the most typical unit of concentration.

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3 years ago

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denpristay [2]

Answer:what digital photo

Explanation:

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4 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago