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GalinKa [24]
3 years ago
7

At what temperature is a gas likely to have the lowest rate of diffusion? 15° 21° 24° 30°

Chemistry
2 answers:
sergiy2304 [10]3 years ago
8 0

Answer:

At 15° the gas will have lowest rate of diffusion.

Explanation:

Temperature directly affects the kinetic energy of gas molecules. The energy of motion is called kinetic energy. So if we increase the temperature of gas the average kinetic energy of the gas molecules will increase (they will move faster). Due to increase in kinetic energy they will easily diffuse from their higher concentration to lower concentration (which is diffusion).

Hence the rate of diffusion at given temperature will have following increasing order.

15° < 21° < 24° < 30°

Ray Of Light [21]3 years ago
7 0

AT 15°

The rate of diffusion of a gas is directly proportional to the absolute temperature. At lower temperature the gas particles have less kinetic energy than at higher temperature. This makes the particles travel faster at higher temperature than lower temperature.

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Vanadium crystallizes in a body-centered cubic lattice, and the length of the edge of a unit cell is 305 pm. What is the density
Anestetic [448]

Answer:

5.96 g/cm^3

Explanation:

Corner atom = 1/8

Atoms in center = 1

Atoms in face of the cube= 1/2  

Molar mass of V = 50.94 g/mol <em>(from period table)</em>

1 mole = 6.02x10^23

<em>In BCC unit cell:</em>

(8 x 1/8)+ 1=2 per 1 unit cell

<em>Mass: </em>2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell

305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)

           = 2.837 x 10^-23 mL

<em>1pm=10^-12m</em>

<em>1cm=10^-2m</em>

<em>1mL=1cm^3</em>

<em></em>

density=mass/volume

density of V = 1.69x10^-22g÷2.837x10^-23mL

                    =5.957g/mL

                    =5.96g/cm^3

5 0
2 years ago
C) Solar energy is the source of all forms of energy.give reasons​
anastassius [24]

Answer:

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2 years ago
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Answer:D.

Explanation:i used photo math. It is really helpful when I'm doing math homework etc...

7 0
3 years ago
If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?
Fiesta28 [93]
The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>).  If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?


(a) Q = K<span>;   The reaction </span>is at equilibrium.
(b) Q < K<span>;   The reaction </span>will proceed to the right.
(c) Q > K<span>;   The reaction </span>will proceed to the left.

The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left, since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
6 0
3 years ago
In which direction does the reaction proceed after heating to 2000 °c?
e-lub [12.9K]
What do I possibly answer here?
ask a full question pls!
6 0
3 years ago
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