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Scilla [17]
1 year ago
5

What is the mass of a 32.5-ml sample of ethanol? the density of ethanol is 0.789 g/ml.

Chemistry
1 answer:
Anon25 [30]1 year ago
6 0

Answer:

25.6 grams of ethanol in 32.5 ml of the substance.

Explanation:

The density of ethanol can be used as a conversion factor.  We know that:

Ethanol is 0.789 g/ml.  That becomes a conversion factor that we may write as (0.789 g/ml) and then convert either grams ethaanol or ml ethanol into the other unit.  We are given volume of ethanol:  32.5 ml.

Note that when we multiply (0.789 g/ml) by (32.5 ml) the ml unit cancels, leaving just grams.  That's what we want, so:

(0.789 g/ml)(32.5 ml) = 25.6 grams of ethanol in 32.5 ml of the substance.

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Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n
lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239

According to Dalton's Law of partial pressure-

P_i=\chi_i\times P_{total}

Where,

P_i=The pressure of the gas component in the mixture

\chi_i= Mole fraction of that gas component

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P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

n=\frac{PV}{RT}

n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole

Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

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4 0
1 year ago
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The equation that relates standard Gibbs free energy, ΔG, with equilibrium constant, K, is:

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<em>Where R is gas constant, 8.314J/molK, and T is absolute temperatue (30.0°C + 273.15 = 303.15K).</em>

<em />

Replacing (110kJ = 110000J):

110000J/mol = -8.314J/molK*303.15K ln K

-43.644 = lnK

1.11x10⁻¹⁹ = K

<em />

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