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2 years ago

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol.

2 answers:

5 0

the balanced chemical equation for the reaction is as follows;
C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

stoichiometry of C₃H₈ to O₂ is 1:5

number of moles of C₃H₈ reacted - 0.025 g / 44.1 g/mol = 0.000567 mol according to molar ratio of 1:5
number of O₂ moles required are 5 times the amount of C₃H₈ moles reacted therefore number of O₂ moles required - 0.000567 x 5 = 0.00284 mol .
mass of O₂ required - 0.00284 mol x 32.00 g/mol = 0.091 mol .
answer is 0.091 mol

Bumek [7]2 years ago

4 0

0.091 is correct on Edgnty!

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Answer : The partial pressure of the hydrogen is, 705.9 mmHg

Explanation :

According to the Dalton's law of partial pressure,

$P_T=P_{H_2}+P_{H_2O}$

where,

$P_T$ = total pressure of the gas = 729.7 mmHg

$P_{H_2}$ = partial pressure of the hydrogen gas = ?

$P_{H_2O}$ = partial pressure of the water = 23.8 mmHg (standard value)

Now put all the given values in the above expression, we get:

$729.7mmHg=P_{H_2}+23.8mmHg$

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3 years ago

1.00 L of gas at a standard temperature and pressure is compressed to 473mL. What is the new pressure of the gas?

Nadusha1986 [10]

Answer:

The pressure of the gas is 2.11 atm.

Explanation:

From the given,

$V_{1}=1.00\,L$

$P_{1}=1\,atm$

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$V_{2}=473\,ml=0.473\,L$

$P_{1}V_{1}=P_{2}V_{2}$

$(1)(1)=(0.473)(P_{2})$

$P_{2}=\frac{(1)(1)}{0.473}=2.11$

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2 years ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

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$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

B) Radius

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

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2 years ago

the development and use of the ______ a common and important piece of laboratory equipment, led scientist to the discovery of se

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The answer is the Microscope.

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3 years ago

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Answer:

It's well Explained below.

Explanation:

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2 years ago