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Reil [10]
3 years ago
15

A gas starts out with a volume of 114 liters and temperature of 273 K. The volume then increases to 216 liters. If the pressure

didn't change, what is the new temperature of the gas?​

Chemistry
1 answer:
Yuri [45]3 years ago
3 0

Answer:

517.3 K

Explanation:

Initial volume of gas V1= 114 L

Initial temp. T1= 273 K

Final volume V2= 216 L

Final temp. T2= ?

From Charles law, Volume is directly proportional to temperature provided pressure is kept constant

V1/T1 = V2/T2

T2 = V2T1/V1

T2 = (216×273)/114

T2 = 517.3 K

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Answer:

D. As white light passes through a prism , it bends and separates into different colors

Explanation:

When light enters a prism, it experiences a phenomenon called 'refraction'.

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The angle at which the ray of light is refracted into the second medium depends on the optical density of the two mediums, but also on the wavelength of the light.

In particular, longer wavelengths (red color) are refracted the least, while shorter wavelengths (violet color) are refracted the most.

As a result, when white light (consisting of all the colors of visible light) enters into a prism, they different wavelengths are separated: therefore, white light separates into different colors.

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3 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
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Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

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A mixture of water and graphite is heated to 600 k in a 1 l container. when the system comes to equilibrium it contains 0.15 mol
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(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???

when the reaction equation is:

C(s) + H2O(g)  ↔ H2(g) + CO(g) 

∴ Kc = [H2] [CO] / [H2O]

and we have Kc = 0.0393 (given missing in the question)

when the O2 is added so, the reaction will be:

2H2(g) + O2(g) → 2H2O(g)

that means that 0.15 mol H2 gives 0.15 mol of H2O

∴ by using ICE table:

            [H2O]          [H2]        [CO]

initial 0.57 + 0.15      0               0.15

change  -X                +X              +X

Equ   (0.72-X)             X            (0.15+X)

by substitution:

0.0393 = X (0.15+X) / (0.72-X)  by solving for X

∴ X = 0.098 

∴[H2] = X = 0.098 M

∴[CO] = 0.15 + X
 
           = 0.15 + 0.098 = 0.248 M

∴[H2O] = 0.72 - X

             = 0.72 - 0.098

             = 0.622 M


8 0
3 years ago
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