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3 years ago

Perform each of the following unit conversions using the conversion factors given below: 1 atm = 760 mmHg = 101.325 kPa

2 answers:

5 0

unit coversation
1.429 atm - 1086mmhg

9361 pa-9.36 KPa - 70.21 mmhg

725 mmhg -0.95 atm- 96.26 kpa

calculation

(a) 1 atm = 760 mmhg
1.429 atm = ?
1.429 x760/1 = 1086.34 mm hg

(B) 1 mmhg = 101.325 kpa
? =9361 KPa
9361 x1 /101.25 =70.21 mmhg

760 mm hg= 101.325 KPa
70.21 mm hg=?

70.21 x101.325/760 = 9.36 Kpa

(C ) 1 atm = 760 mmhg
? = 725
= 725 x1/ 760=0.95 atm

1 atm = 101.325 kpa
0.95 =?

0.95 x101.325/1 = 96.26 KPa

Leto [7]3 years ago

4 0

1atm= 760mmhg= 101.325kPa

1.429atm= 1,086 mmhg

9,361 Pa= 9.361 kPa= 70.21 mmhg

725 mmhg= 954 atm= 96.7 kPa

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Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

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3 years ago