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3 years ago

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explain why the sun appers to be the brightest star in the sky. relate your explanation with an example such as street lights. t

his is science

1 answer:

Zarrin [17]3 years ago

7 0

Answer:

The sun is a star but It is the brightest in the sky because it is the closest star to earth.

I don't know how to compare it to street lights though...

Explanation:

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$

The mole fraction of nitrogen in the mixture is:

$X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33$

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

5 0

3 years ago

A sample of nitrogen gas occupies a volume of 1.40L at 297K with a pressure of 776mmHg. The volume increases to 2.70L and the te

konstantin123 [22]

Hey it isnt letting me submit my answer to your question on the Japanese chart, so imma just submit it here

6 0

3 years ago

Convert the following to scientific notation 24,212,000

UNO [17]

2.4212 X 10^ 7
How I at least figure this problem out is I take a pencil and start on the right side of the 0 and make a loop to the left for each number and count until I get to the first two numbers that are between 1-9 when reading from left to right. This is where you put the decimal point. Some teachers rather you keep the 0's there, while others prefer one to get rid of them. Anyways with that new decimal number, you multiply the decimal by ten to what ever number you counted, which was 7.

5 0

3 years ago

What is the definition of Axis of rotation

il63 [147K]

Answer: an invisible line around which an object rotates, or spins.

Explanation: //Give thanks(and or Brainliest) if helpful (≧▽≦)//

6 0

3 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

3 years ago