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3 years ago

Which of the following is a combustion reaction?

Chemistry

2 answers:

Zielflug [23.3K]3 years ago

8 0

Answer:

B. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Explanation:

A combustion reaction can be defined as a class of chemical reactions, that named as "burning", that occurs usually when a <em>hydrocarbon reacts with oxygen to produce carbon dioxide (CO₂) and water (H₂O).</em>

So, the correct answer is the choice B, 1.0 mole of propane "hydrocarbon" burned with 5.0 moles of oxygen to produce 3.0 moles of CO₂ and 4.0 moles of H₂O.

snow_lady [41]3 years ago

5 0

That would be B.

Its the combustion of propane in oxygen.

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How many grams of NaCl are in 0.78 moles of NaCl? *

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3 years ago

What mass of Al contains the same number of atoms as 33.5g of Au

wel

Stoich questions...

1) convert the grams of Au to moles
(33.5g)/(196.7g/mol)=___x___

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x(moles of Au)*(6.02*10^23)=?

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3 years ago

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Lelu [443]

Answer:

Solar eclipses

Explanation:

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3 years ago

Concerning Boyle's Law, if you had a gas at a pressure of 101 kPa and decreased the volume of the container holding the gas to o

Viefleur [7K]

Answer:

$P_2=404 kPa$

Explanation:

Hello,

In this case, the Boyle's is mathematically defined via:

$P_1V_1=P_2V_2$

Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:

$V_2=\frac{1}{4} V_1$

We can compute the new pressure:

$P_2=\frac{P_1V_1}{V_2} =\frac{P_1V_1}{\frac{1}{4} V_1} =\frac{101kPa*V_1}{\frac{1}{4} V_1} \\\\P_2=4*101kPa\\\\\\P_2=404 kPa$

Which means the pressure is increased by a factor of four.

Regards.

7 0

3 years ago