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professor190 [17]
4 years ago
10

Please help pleaseeee

Chemistry
1 answer:
mylen [45]4 years ago
8 0

Answer:

The answer is C) two covalent bonds between adjacent molecules.

Explanation:

The exact definition of a double bond is, "a chemical bond in which two pairs of electrons are shared between two atoms. "<em> - Taken from Google Dictionary</em>

<em />

<em>The following information was taken from Wiki to help you understand the question, and so that no one thinks that I plagiarized: </em>

"A double bond in chemistry is a chemical bond between two chemical elements involving four bonding electrons instead of the usual two. The most common double bond occurs between two carbon atoms and can be found in alkenes. Many types of double bonds exist between two different elements."

I hope this helps!

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>
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7 0
3 years ago
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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

3 0
3 years ago
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mrs_skeptik [129]

Answer:

16

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4 years ago
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Answer:

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