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2 years ago

What occurs when a magnesium atom becomes a magnesium ion

Chemistry

2 answers:

Law Incorporation [45]2 years ago

7 0

Answer is: Electrons are lost and the oxidation number increases.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium atom: ₁₂Mg 1s² 2s² 2p⁶ 3s².

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Karo-lina-s [1.5K]2 years ago

4 0

The electrons are lost.

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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

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

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

Which of these lists the energies of the sublevels, from highest to lowest? A. f > d > p > s B. d > p > s > f

Pie

A because for example in n=4 we have all of these sub levels 4s is completed sooner than 4p and so

4 0

3 years ago

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

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

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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3 years ago

26 g of zinc combines with 12.8 g of sulfur. What is the empirical formula of zinc sulfide?

Alisiya [41]

Moles of Zn: 26 / 65 = 0.4
Moles of S: 12.8 / 32 = 0.4

Molar ratio of Zn : S = 1 : 1
Empircal formula: ZnS

The answer is C

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3 years ago

How many grams oxygen are present in 76.8g of CO2

katrin [286]

There would be 55.8 g present

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2 years ago

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