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Usimov [2.4K]
3 years ago
8

when sodium atom reacts with chlorine atoms form a compound, the electron configurations of ions forming the compound are the sa

me as those in which noble gas atoms
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
A sodium atom has the electron configuration of [ 2,8,1 ] before bonding with a chlorine atom which has the electron configuration of [ 2,8,7 ].

because sodium becomes a one plus ion ( +1 ), it must lose an electron to become stable therefore when bonded the electron configuration becomes [ 2,8 ] this is the same as the noble gas, Neon [ 2,8 ].

because chlorine becomes a minus one ion ( -1 ), it must gain an electron to become stable therefore when boded the electron configuration becomes [ 2,8,8 ] this is the same as the noble gas, Argon [ 2,8,8 ].

hope this helps :)
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9.54 x 10^-4 mol/L is the solubility of lead(II) iodide in water at 0 C. What is the solubility product constant for lead(II) io
chubhunter [2.5K]

Answer:

Ksp = 3.47x10⁻⁹

Explanation:

When Lead (II) iodide (PbI₂) is added to water, the equilbrium produced is:

PbI₂(s) → Pb²⁺(aq) + 2 I⁻(aq)

And solubility product constant, ksp is:

Ksp = [Pb²⁺] [I⁻]²

A solubility of 9.54x10⁻⁴ M means the maximum concentration of Pb²⁺ is 9.54x10⁻⁴ M and 9.54x10⁻⁴ M×2 of I⁻. Replacing in ksp formula:

Ksp = [9.54x10⁻⁴] [2×9.54x10⁻⁴]²

<em>Ksp = 3.47x10⁻⁹</em>

5 0
3 years ago
The major reason why relative atomic mass of some elements is not a whole number the existence​
LuckyWell [14K]

Answer:

Because of the existence of isotopes.

4 0
3 years ago
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How would decreasing the pressure affect the equilibrium of this reaction? Hat 12 32H1 O A. The reaction would remain in equilib
Harrizon [31]

A. The reaction would remain in equilibrium

<h3>Further explanation</h3>

Given

Reaction

H₂ + I₂ ⇔ 2HI

Required

the effect of pressure changes

Solution

In the equilibrium system :

<em>Reaction = - action   </em>

⇒shift the reaction to the right or left.  

The pressure usually affects the gas equilibrium system(only count the number of moles of gases)

The addition of pressure, the reaction will shift towards a smaller reaction coefficient  ((the fewest moles of gas  )

Reaction

H₂ + I₂ ⇔ 2HI

The reactant side of the equation has 2 moles of a gas(1 mole H₂ and 1 mole  I₂) ; the product side has 2 moles HI

So the total number of moles from both sides is the same(2 moles) so that the change in volume (pressure) <em>does not change the direction of equilibrium⇒No shift will occur </em>

6 0
3 years ago
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solmaris [256]
The shape of the H2O molecule is a Bent Triatomic.
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6 0
3 years ago
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Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the sta
Hatshy [7]

<u>Answer:</u> The pressure equilibrium constant for the reaction is 1473.8

<u>Explanation:</u>

We are given:

Initial partial pressure of methane gas = 2.4 atm

Initial partial pressure of water vapor = 3.9 atm

Equilibrium partial pressure of hydrogen gas = 6.5 atm

The chemical equation for the reaction of methane gas and water vapor follows:

                        CH_4+H_2O\rightleftharpoons CO+3H_2

<u>Initial:</u>               2.4       3.9

<u>At eqllm:</u>        2.4-x    3.9-x        x       3x

Evaluating the value of 'x':

\Rightarrow 3x=6.5\\\\x=2.167

So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

Equilibrium partial pressure of water vapor = (3.9 - x) = [3.9 - 2.167] = 1.733 atm

Equilibrium partial pressure of carbon monoxide gas = x = 2.167 atm

The expression of K_p for above equation follows:

K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}

Putting values in above equation, we get:

K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8

Hence, the pressure equilibrium constant for the reaction is 1473.8

5 0
3 years ago
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