Which statement is true about a neutral solution?

consider the balanced chemical reaction below and determine the percent yield for carbon dioxide if 4.5 moles of propane yielded

valina [46]

What is the formula for the question

6 0

3 years ago

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if the mass of the logs were 10kg before the fire, what would the mass of the ash, carbon dioxide and water be after the fire

blsea [12.9K]

Answer:

10kg

Explanation:

because matter can not be created or destroyed

4 0

3 years ago

Lesson Question: What is the effect of

tangare [24]

Answer: what is the effect of ( pressure ) on the volume of a gas

If the pressure on a gas is increased, then it’s volume will ( decrease ) because the gas molecules will ( be pushed closer together)

Explanation:

8 0

3 years ago

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A gas that occupies 50.0 liters has its volume increased to 68 liters when the pressure was changed to 3.0 ATM. What was the ori

Yuliya22 [10]

Answer:

4.1 atm = 3,116 mmHg = 415.4 kPa

Explanation:

According to Boyle's law, as volume is increased the pressure of the gas is decreased. That can be expressed as:

P₁ x V₁= P₂ x V₂

Where P₁ and V₁ are the initial pressure and volume respectively, and P₂ and V₂ are final pressure and volume, respectively.

From the problem, we have:

V₁= 50.0 L

V₂= 68.0 L

P₂= 3.0 atm

Thus, we calculate the initial pressure as follows:

P₁= (P₂ x V₂)/V₁= (3.0 atm x 68.0 L)/(50.0 L)= 4.08 atm ≅ 4.1 atm

To transform to mmHg, we know that 1 atm= 760 mmHg:

4.1 atm x 760 mmHg/1 atm = 3,116 mmHg

To transform to kPa we use: 1 atm= 101.325 kPa

4.1 atm x 101.325 kPa = 415.4 kPa

5 0

3 years ago

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31

Anastaziya [24]

Answer:

The volume of the gas will be 78.31 L at 1.7 °C.

Explanation:

We can find the temperature of the gas by the ideal gas law equation:

$PV = nRT$

Where:

n: is the number of moles

V: is the volume

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

From the initial we can find the number of moles:

$n = \frac{P_{1}V_{1}}{RT_{1}} = \frac{1 atm*62.65 L}{(0.082 L*atm/K*mol)*(0 + 273)K} = 2.80 moles$

Now, we can find the temperature with the final conditions:

$T_{2} = \frac{P_{2}V_{2}}{nR} = \frac{612.0 mmHg*\frac{1 atm}{760 mmHg}*78.31 L}{2.80 moles*0.082 L*atm/(K*mol)} = 274.7 K$

The temperature in Celsius is:

$T_{2} = 274.7 - 273 = 1.7 ^{\circ} C$

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.

I hope it helps you!

8 0

3 years ago