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Svetach [21]
3 years ago
12

A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.

Chemistry
1 answer:
Katarina [22]3 years ago
6 0

Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

R= 0.0821Latm/molK

T = 158 C = 158 +273 K = 431 K

the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

n = 4.87 * 67.54 / 0.0821 * 431

n = 328.9198 / 35.3851

n = 9.295moles

The number of moles is approximately 9.30moles.

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How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?
vova2212 [387]

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

7 0
3 years ago
Combining Boyle's law and Charles' law forms one presure-temperature-volume relationship and holds the mass of the gas constant.
Oksanka [162]

(P1V1/T1)=(P2V2/T2)

I just took the test and this should be correct

4 0
3 years ago
Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneo
Mashutka [201]

Answer:

a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance

Explanation:

Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.

8 0
3 years ago
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0
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Which alkane may be obtained by the reduction of 2-bromo propane
Sunny_sXe [5.5K]

To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.

8 0
3 years ago
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