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2 years ago

You have a sample of 75.6 g of C3H8. How many moles of C3H8 are in the sample?

Chemistry

1 answer:

dsp732 years ago

3 0

The number of moles that are contained in the given mass of propane ( $C_3H_8$ is 1.7143 moles.

<u>Given the following data:</u>

Mass of propane = 75.6 grams.

<u>Scientific data:</u>

The molar mass of propane = 44.1 g/mol.

To calculate the number of moles that are contained in the given mass of propane ( $C_3H_8$ ):

<h3>How to calculate the moles of a compound.</h3>

In this exercise, you're required to determine the number of moles of propane that are contained in the given sample:

Mathematically, the number of moles contained in a chemical compound is given by this formula:

$Number\;of\;moles = \frac{mass}{molar\;mass}$

Substituting the given parameters into the formula, we have;

$Number\;of\;moles = \frac{75.6}{44.1}$

Number of moles = 1.7143 moles.

Read more on number of moles here: brainly.com/question/3173452

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When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu

expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water = 1000J

Mass of water = 50g

Final temperature = 30°C

Unknown:

Initial temperature = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

1000 = 50 x 4.184 x (30 - initial temperature )

1000 = 209.2(30 - initial temperature)

4.78 = 30 - initial temperature

4.78 - 30 = - initial temperature

Initial temperature = 25.2°C

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2 years ago

A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3

ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

$V_1=250ml$

$V_2=?$

$P_{total}=735 mmhg$

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

$P_{water}=42.2mmhg$

$T_1=35°C=35+273=308 K$

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

$P_{gas}=P_{total}-P_{water}$

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

$\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}$

$V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1$

$V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1$

Here $P_2 \:and\: T_2$ are standard pressure and temperature respectively.

we have

$P_2=750mmhg \:and\: T_2=273K$

Substituting value, we get

$V_2=\frac{692.8}{750}*\frac{273}{308} *250$

$V_2= 819.51 ml$

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1 year ago

What is the correct name for SO3?

ElenaW [278]

Sulfur Trioxide is the correct name for SO3

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2 years ago

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Vikki [24]

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Explanation:

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3 years ago

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Marina86 [1]

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2 years ago