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Ilya [14]
3 years ago
5

You transfer a sample of a gas at 17°C from a volume of 5.67 L and 1.10 atm to a container at 37°C that has a pressure of 1.10 a

tm. What is the new volume of the gas?
Chemistry
1 answer:
laila [671]3 years ago
5 0

Explanation:

V1T1 = V2T2

V2=V1T1/T2

= 5.67*17/37= 2.6

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A compound with the empirical formula CH2 was found to have a molar mass of approximately 112 g. Write the molecular formula of
san4es73 [151]

Answer:

C8H10

Explanation:

n (CH2) = 112

n (12 + 1 + 1) = 112

n (14) = 112

n = 8

Molecular Formula: C8H10

6 0
2 years ago
Please select the word from the list that best fits the definition
leonid [27]

Answer:

Systemic Circulation

Explanation:

7 0
2 years ago
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A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
What happens when molecules diffuse across the cell membrane?
mafiozo [28]

Answer:

The correct answer is from areas of high concentration to low concentration.

Explanation:

A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.

Hope this Helps!

7 0
2 years ago
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The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

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At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
3 years ago
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