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2 years ago

The highest energy occupied molecular orbital in the li−li bond of the li2 molecule is _____.

Chemistry

2 answers:

sveta [45]2 years ago

6 0

Answer:

Explanation:

Hello,

To determine the stated orbital, we must develop the electron configuration of Li:

$1s^22s^1$

So the highest energy occupied molecular orbital is 2s which is the most stable condition.

Best regards.

Lynna [10]2 years ago

4 0

The answer is toluene I think

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Sam buys a package of chips and notices that hydrogenated vegetables oil is one of the ingredients. This ingredient primarily:

Vladimir [108]

Hydrogenated vegetable oil is a preservative, thus, the correct option is D. Hydrogenated vegetable oil are used by food industries to prevent fat rancidity in order to give their product a longer shelf life. Hydrogenated oils are very good preservatives because all the enzymatic activities in the oil has been neutralized during the hydrogenating process.

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2 years ago

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

Dennis_Churaev [7]

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

Volume of the calcium hydroxide solution used is 0.0235 mL.

4 0

2 years ago

Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

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

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

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago

BRAINLIESTTT ASAP!!!

Stels [109]

While I am not the brainliest I can certainly answer.

This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.

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Luda [366]

Answer:

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Explanation:

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2 years ago

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