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jok3333 [9.3K]

The bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Since the chemical reaction is 2CO + O₂ → 2CO₂ and the total bond energy of the products carbon dioxide CO₂ is 1,472 kJ.

Since from the chemical reaction, we have 2 moles of CO₂ which gives 1,472 kJ and there are two carbon-oxygen, C-O bonds in CO₂, then

2 × C-O bond = 1,472 kJ

1 C-O bond = 1.472 kJ/2

C-O bond = 736 kJ

So, the bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Learn more about bond energy here:

brainly.com/question/21670527

7 0

2 years ago

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A substance is matter that has a definite composition.<br> TRUE or FALSE?

Greeley [361]

Answer: false

Explanation:

3 0

3 years ago

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Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

2 years ago

Ethene is the first member of the

DanielleElmas [232]

It will be:

b. Alkyne* series

* You have done a spelling mistake here. It will be "Alkene" not "Alkyne".

5 0

3 years ago

To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}$

7 0

3 years ago