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kicyunya [14]
3 years ago
7

Why is the? The second ionization energy of sodium is about three times greater than the second ionization energy of magnesium

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

~Na+1 is already in the preferred form. Because of this, the second ionization energy of sodium is higher than normal. Mg+1 loses an electron to form s2 p6 .

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What is biological function?
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Answer:

Biological Functions: processes and structures by which organisms adjust to short-term or long-term changes in their environment

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2 years ago
Which of the following explains the characteristic melting and boiling points of ionic compounds? in ionic solids, the ions are
Wewaii [24]
FOLLOWING!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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4 years ago
The standard enthalpy change for the following reaction is 873 kJ at 298 K.
Flura [38]

Answer:  - 436.5 kJ.

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reaction is,

2KCl(s)\rightarrow 2K(s)+Cl_2(g)  \Delta H_1=873kJ

Now we have to determine the value of \Delta H for the following reaction i.e,

K(s)+\frac{1}{2}Cl_2(g)\rightarrow KCl(s) \Delta H_2=?

According to the Hess’s law, if we divide the reaction by half then the \Delta H will also get halved and on reversing the reaction , the sign of enthlapy changes.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=\frac{1}{2}\times (-873kJ)

\Delta H_2=-436.5kJ

Hence, the value of \Delta H_2 for the reaction is -436.5 kJ.

8 0
3 years ago
Which of these is an element?
just olya [345]
B
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3 0
4 years ago
Read 2 more answers
Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat
larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

ΔG° = -377.71 kJ = -3.7771 × 10² kJ

This is the standard Gibbs free energy per mole of reaction.

5 0
3 years ago
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