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Molodets [167]
3 years ago
5

Which of the following is not a valid equation for describing the behavior of

Chemistry
2 answers:
bonufazy [111]3 years ago
5 0
D is the answer!!!!!!!!!!!!!!!!
AnnZ [28]3 years ago
3 0

Answer:

A. (P1/V1) = (P2/V2) is not a valid equation

D. (V2/T1) =V2/T2) is not a valid equation.  ( V1/T1) = (V2/T2) is a valid equation

Explanation:

Step 1: Data given

The gas law is p*V = nRT (this means <u>option B is a valid equatio</u>n)

If we compare 2 gases we can write this as:

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )

Step 2: When temperature and number of moles are constant

This means volume and pressure will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )  OR p1¨V1 = p2*V2

The relationship for Boyle’s Law can be expressed as follows: p1V1 = p2V2, where p1 and V1 are the initial pressure and volume values, and p2 and V2 are the changed values of the pressure and volume of the gas.

When the initial pressure rises, the initial volume will decrease (and vice versa). For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.

For the formula (P1/V1) = (P2/V2) it would mean pressure and volume are directly proportional, this is not true. So<u> option A is not a valid equation. </u>

Step 3: When pressure and number of moles are constant

This means volume and temperature will change.

p1*V1 / ( n1*R*T1 ) = p2*V2 / ( n2*R*T2 )  OR V1/T1 = V2/T2

For the formula (V1/T1) =V2/T2) it would mean volume  and temperature are directly proportional. This<u> is a valid equation</u>

<u>Option D</u> would only be a valid equation if its (V1/T1) =V2/T2)

(V2/T1) =V2/T2) is not a valid equation. ( I don't know if this was a thypo or not).

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Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

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c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

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A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu
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2) The Kelvin degree is written as K, not ºK

Explanation:

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Therefore  in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

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The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate
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<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

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