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3 years ago

Please help! Don’t know if I’m right!!

Chemistry

1 answer:

kompoz [17]3 years ago

4 0

Answer:e

Explanation:

Bananas

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Isobutane (C4H10) is one of the components of natural gas. Which equation

Gnom [1K]

Answer:

D. C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O + heat

Explanation:

The reaction of the combustion of isobutane with oxygen is presented as follows;

C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O + heat

In the above reaction, the number of elements on the reactant side are;

4 carbon, C, atoms, 10 hydrogen, H, atoms, and 13 oxygen, O, atoms

The number of elements on the product side of the above chemical reaction are;

The number of carbon atoms, C = 4

The number of hydrogen atoms, H = 10

The number of oxygen atoms, O = 13

Therefore, the number of atoms on the reactant side of the chemical equation are equal to the number of atoms on the product side, and the combustion reaction for isobutane is balanced.

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3 years ago

Write a balanced nuclear equation for the beta decay of iron-59

ruslelena [56]

26Fe59 ----> -1e0 + 27Co59

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2 years ago

When a drawing made with a black marker gets wet, the marker bleeds and more separates into several colors. What method of separ

Mnenie [13.5K]

This example is describing the Chromatography method.

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3 years ago

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Write down the structural formula for the monomer from which

Sati [7]

Answer:

Vinyl chloride is an organochloride with the formula H2C=CHCl that is also called vinyl chloride monomer (VCM) or chloroethene. This colorless compound is an important industrial chemical chiefly used to produce the polymer polyvinyl chloride (PVC).

Explanation:

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2 years ago

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How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?

nika2105 [10]

Answer:

$m_{C_8H_{18}}=85.67gC_8H_{18}$

Explanation:

Hello there!

In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

$m_{C_8H_{18}}=300.0gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molC_8H_{18}}{25molO_2}*\frac{114.22gC_8H_{18}}{1molC_8H_{18}} \\\\m_{C_8H_{18}}=85.67gC_8H_{18}$

Regards!

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2 years ago