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Alla [95]
3 years ago
9

Write and balance the combustion equation for propane. (Propane is combusted in the presence of oxygen to produce carbon dioxide

and water). 2) How many grams of oxygen are required to burn 200 grams of propane
Chemistry
2 answers:
disa [49]3 years ago
6 0

Answer:

363.64g of oxygen would be required.

Explanation:

1) Write and balance the combustion equation for propane.

Propane + Oxygen --> carbon dioxide + water

C3H8 + O2 --> CO2 + H2O

Upon balancing, we have;

C3H8 + 5O2 --> 3CO2 + 4H2O

2) How many grams of oxygen are required to burn 200 grams of propane.

From the reaction;

Propane = (3 * 12) + (8 * 1) = 44

Oxygen = (5 * 16) = 80

80 grams of oxygen is required to combust 44g of propane.

80 = 44

x = 200

x = ( 80 * 200 ) / 44

x = 363.64g

olga nikolaevna [1]3 years ago
4 0

Answer:

C3H8 + 5O2 → 3CO2 + 4H2O

We need 725.6 grams O2 to burn 200 grams of propane

Explanation:

Step 1: Data given

propane = C3H8

Molar mass propane = 44.1 g/mol

Mass of propane = 200.0 grams

Molar mass of O2 = 32.0 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 3: Calculate moles propane

Moles propane = mass propane / molar mass propane

Moles propane = 200.0 grams / 44.1 g/mol

Moles propane = 4.535 moles

Step 4: Calculate moles O2

For 1 mol C3H8 we need 5 moles O2 to react to produce 3 moles CO2 and 4 moles H2O

For 4.535 moles propane we need 5*4.535 = 22.675  moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 22.675 moles * 32.0 g/mol

Mass O2 = 725.6 grams

We need 725.6 grams O2 to burn 200 grams of propane

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In the partition coefficient experiment 4A this week, after thorough mixing of the reagents, phase separation will occur. The to
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Answer:

The top layer is the Aqueous layer, and the benzoic acid is contained in the non-aqueous layer/oil phase.

Explanation:

A separating funnel is a very important piece of laboratory glassware that is used to separate the components of liquid-liquid mixtures  which are immiscible. This technique is used in the extraction of the components of mixtures.

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A mixture of helium, nitrogen and oxygen has a total pressure of 756 mmHg. The partial
Karolina [17]

Answer:

The  partial pressure of oxygen in the mixture is 296 mmHg.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone.

This relationship is due to the assumption that there are no attractive forces between the gases.

So, in this case, the total pressure is:

PT=Phelium + Pnitrogen + Poxygen

You know:

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Replacing:

756 mmHg= 122 mmHg + 338 mmHg + Poxygen

Solving:

756 mmHg - 122 mmHg - 338 mmHg = Poxygen

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<u><em>The  partial pressure of oxygen in the mixture is 296 mmHg.</em></u>

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3 years ago
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate
Vika [28.1K]

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

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You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

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If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

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We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

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