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Nina [5.8K]
3 years ago
6

Propane is used as a fuel for camp stoves. It undergoes combustion to form carbon dioxide and water.

Chemistry
1 answer:
Arisa [49]3 years ago
8 0
Fuel camp equal width time inchess
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What is the structure of beeswax that contains 16 carbon of carboxylic acid and 30 carbon of alcohol?
melisa1 [442]
This is the answer hope this helps

7 0
3 years ago
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What is the volume of 8.80 g of CH4 gas at STP?
Ann [662]

Answer:

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.

6 0
3 years ago
In the compound H2O, the electrons in the bonds are unequally shared between oxygen and hydrogen, forming ____. (1 point)
Aloiza [94]
<span>polar bonds due to high difference . In electronegativity of oxygen and hydrogen.</span>
5 0
3 years ago
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Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
The air all around us is a mixture of gasses containing nitrogen, oxygen, carbon dioxide, argon
KiRa [710]

Answer:

Total pressure = 4.57 atm

Explanation:

Given data:

Partial pressure of nitrogen = 1.3 atm

Partial pressure of oxygen = 1824 mmHg

Partial pressure of carbon dioxide = 247 torr

Partial pressure of argon = 0.015 atm

Partial pressure of water vapor = 53.69 kpa

Total pressure = ?

Solution:

First of all we convert the units other into atm.

Partial pressure of oxygen = 1824 mmHg / 760 = 2.4 atm

Partial pressure of carbon dioxide = 247 torr / 760 = 0.325 atm

Partial pressure of water vapor = 53.69 kpa / 101 = 0.53 atm

Total pressure = Partial pressure of N +  Partial pressure of O +  Partial pressure of CO₂ +  Partial pressure of Ar +  Partial pressure of water vapor

Total pressure = 1.3 atm + 2.4 atm + 0.325 atm + 0.015 atm + 0.53 atm

Total pressure = 4.57 atm

4 0
3 years ago
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