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3 years ago

Cobalt (I) fluoride how do i write this into a formula?

Chemistry

1 answer:

Nuetrik [128]3 years ago

5 0

Answer:

The answer would be CoF for Cobalt (I) Fluoride.

Explanation:

So first of all you take your periodic table and look at the valence electrons of Fluorine...it is 1. You don't need to look the valence electron of Cobalt because the "I" after Cobalt tells its valence electrons (it is a transition metal that is why). Now that Cobalt has a +1 charge and Fluorine has a -1 charge, it is automatically nuclear (means charge are cancelled out to 0). Therfore the answer is CoF.

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Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

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. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

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Answer:

Explanation:

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The formula for the change in Gibbs energy of a solid is:

ΔG = Vm ΔP

where, ΔG is change in Gibbs, Vm is molar volume, ΔP is change in pressure

ΔP = P(final) – P(initial)

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So,

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Explanation:

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