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bogdanovich [222]
3 years ago
15

Butane (c4h10) is liquid fuel used in lighters. how many grams of carbon are in a lighter containing 7.25 mL of butane? The dens

ity of liquid butane is 0.601 g/mL
Chemistry
1 answer:
Alisiya [41]3 years ago
6 0
Density = 0.601 g/mL

Volume = 7.25 mL

D = m / V

0.601 = m / 7.25

m = 0.601 x 7.25 => 4.35725 g

hope this helps!
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While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by
grin007 [14]

Answer:

n_{C2H_5OH}^{eq}=14.234mol

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
What volume did a helium-filled balloon have at 22.5 c and 1.95 atm if it’s new volume was 56.4 mL at 3.69 atm and 11.9c
Veseljchak [2.6K]

This is an exercise in the general or combined gas law.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • T₁ = 22.5 °C + 273 = 295.5 K
  • P₁ = 1.95 atm
  • V₁ = ¿?
  • P₂ = 3.69 atm
  • T₂ = 11.9 °C + 273 =  284.9 k
  • V₂= 56.4 ml

We use the following formula:

P₁V₁T₂ = P₂V₂T₁ ⇒ General formula

Where

  • P₁ = Initial pressure
  • V₁ = Initial volume
  • T₂ = Initial temperature
  • P₂ = Final pressure
  • V₂ = final volume
  • T₁ = Initial temperature

We clear the formula for the initial volume:

\boldsymbol{\sf{V_{1}=\dfrac{P_{2}V_{2}T_{1}}{P_{1}T_{2}}  } }

We substitute our data into the formula to solve:

\boldsymbol{\sf{V_{1}=\dfrac{(3.69 \not{atm})(56.4 \ ml)(295.5 \not{k})}{(1.95 \not{atm})(284.9\not{k})}  }}

\boldsymbol{\sf{V_{1}=\dfrac{61498.278}{555.555} \ lm }}

\boxed{\boldsymbol{\sf{V_{1}=110.697 \ lm }}}

The helium-filled balloon has a volume of <u>110.697 ml.</u>

6 0
1 year ago
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