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3 years ago

From which sublevel are electrons removed when a zn atom in the ground state is oxidized?

Chemistry

1 answer:

Vesnalui [34]3 years ago

5 0

Answer:

4s2

Explanation:

The electronic configuration of zinc in the ground state of the atom is 1s22s22p6 3s23p63d104s2. It can be written in a condensed form by showing the inert gas core as follows; [Ar] 3d¹⁰ 4s².

Zinc has a pseudo inert gas configuration when the 4s electrons are removed during oxidation. Hence this sublevel is usually affected when zinc atoms are oxidized according to the scheme;

Zn---->Zn^2+ + 2e-

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2 years ago

A 2.0 M solution of LiF had 4.00 moles of LiF added to a solvent to make it.

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Answer:

2 L

Explanation:

From the question given above, the following data were obtained:

Molarity of LiF = 2 M

Mole of LiF = 4 moles

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Molarity of a solution is simply defined as the mole per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole / Volume

With the above formula, we can obtain the volume of the solution as shown below:

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2 years ago

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3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$