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goldenfox [79]
3 years ago
7

What is the volume of a 3.0 M solution of hydrochloric acid that contains 1.50 moles of solute?

Chemistry
1 answer:
Nuetrik [128]3 years ago
6 0
C=3.0 mol/L
n=1.50 mol

n=cv

v=n/c

v=1.50/3.0=0.5 L = 500 mL

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snow_tiger [21]
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Katyanochek1 [597]

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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3 years ago
True or False
lozanna [386]
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3 years ago
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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3 years ago
When a bullet is fired from a gun what is true about the momentum
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Before the bullet is fired the momentum is Zero because nothing is moving but once the bullet is shot the momentum increases because of the movement of the bullet moving forward.

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