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3 years ago

How many valence electrons are transferred from the calcium atom to iodine in the formation of the compound calcium iodide?

Chemistry

1 answer:

Andreas93 [3]3 years ago

7 0

Answer:

Two electrons.

Explanation:

Calcium is the element of second group and forth period. The electronic configuration of Calcium is - 2, 8, 8, 2 or $1s^22s^22p^63s^23p^64s^2$

There are 2 valence electrons of Calcium.

Iodine is the element of seventeenth group and fifth period. The electronic configuration of Iodine is - 2, 8, 18, 18, 7 or $[Kr]4d^{10}5s^25p^5$

There are 7 valence electrons of iodine.

Calcium iodide, $CaI_2$ is formed when 2 valence electrons of calcium are loosed and they are gained by 2 atoms of iodine.

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Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

2 years ago

The reaction below virtually goes to completion because cyanide ion forms very stable complexes with Ni2+ ion:[Ni(H2O)6]2+(aq) +

MAVERICK [17]

Answer:

Option A and D are correct.

Unstable species react rapidly.

Stable species do not react rapidly.

Explanation:

The complete question is attached to this solution.

The more stable a reactant is, the less reactive it will be. A stable reactant has a very stable structure in which it will avoid any perturbations. And for a reaction to occur, the bonds in the reactant must break down to form the products. A stable reactant has very strong bonds that aren't easy to break down, hence, reactions involving very stable reactants do not proceed rapidly.

And the more unstable a reactant specie is, the more rapidly it reacts. This is why the reaction involving the less stable isotope of carbon; Carbon-14 is very rapid. It is the same reason as explained above that is responsible for this. The bond between unstable species are not strong and are easily breakable, thereby leading to a quick reaction.

Hope this Helps!!!

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