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nekit [7.7K]
3 years ago
12

 

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

The answer to your question is     V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L                    Volume 2 =

Temperature 1 = T1 = 19°C               Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg           Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

                  P1V1/T1  = P2V2/T2

-Solve for V2

                  V2 = P1V1T2 / T1P2

-Substitution

                  V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

                 V2 = 1068012 / 214620

-Result

                 V2 = 4.97 l

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What is difference between ∆H and ∆H°
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Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of
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  • Density=0.71g/ml

\boxed{\sf Density=\dfrac{Mass}{Volume}}

\\ \sf{:}\implies Mass=Density(Volume)

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6 0
2 years ago
Be sure to answer all parts. find the molar solubility of bacro4 (ksp= 2.1 × 10−10) in (a) pure water × 10 m (b) 1.6 × 10−3 m na
Vlad [161]
A) in pure water :

by using ICE table:

According to the reaction equation:

            BaCrO4(s)    →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                               0                          0

change                          +X                       +X 

Equ                                  X                         X


when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X* X

∴X = √2.1 x 10*-10

∴X = 1.4 x 10^-5

∴ the solubility = X = 1.4 X 10^-5

B) In 1.6 x 10^-3 m Na2CrO4

 by using ICE table:

According to the reaction equation:

            BaCrO4(s)  →  Ba^2+(aq)    +   CrO4^2-(aq)

initial                                 0                      0.0016

Change                           +X                      +X

Equ                                   X                      X+0.0016

when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X*(X+0.0016) by solving for X 

∴ X = 1.3 x 10^-7

∴ solubility =X = 1.3 x 10^-7

3 0
3 years ago
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