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SOVA2 [1]
3 years ago
7

For a particular isomer of C8H18, the combustion reaction produces 5104.1 kJ of heat per mole of C8H18(g) consumed, under standa

rd conditions. C8H18(g)+252O2(g)⟶8CO2(g)+9H2O(g)ΔH∘rxn=−5104.1 kJ/mol What is the standard enthalpy of formation of this isomer of C8H18(g)?
Chemistry
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

Explanation:

Hello,

Considering the chemical reaction, the enthalpy of reaction is given by:

ΔH°rxn=ΔfHCO2+ΔfHH2O-ΔfHC8H18

(ΔfHO2=0)

Taking into account that the reaction produces energy, ΔH°rxn is negative. No, solving for ΔfHC8H18:

ΔfHC8H18=-ΔH°rxn+8*ΔfHCO2+9*ΔfHH2O

ΔfHC8H18=-(-5104.1 kJ/mol)+9*(-292.74kJ/mol)+8*(-393.5 kJ/mol)

ΔfHC8H18=-678.56 kJ/mol

Best regards.

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it is an endothermic because water can not give off heat it can only take from others like if you were to boil water the water is endothermic and the heat is the exothermic

3 0
3 years ago
A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
Mazyrski [523]

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

PV_1=nRT

where,

P = pressure of gas = 1 atm

V_1 = volume of gas = ?

T = temperature of gas = 270^oC=273+270=543K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K

V_1=9.12L

Now we have to calculate the volume at 24°C.

PV_2=nRT

where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 9.12 L

V_2 = final volume = 4.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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3 years ago
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3 years ago
A Helium gas in a tube with a volume of 9.583 L under pressure of 4.972 atm at 31.8 c
andre [41]

1.905 moles of Helium gas are in the tube. Hence, option A is correct.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 4.972 atm

V= 9.583 L

n=?

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=31.8 +273= 304.8 K

Putting value in the given equation:

\frac{PV}{RT}=n

n= \frac{4.972 \;atm\; X \;9.583 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 304.8}

Moles = 1.905 moles

1.905 moles of Helium gas are in the tube. Hence, option A is correct.

Learn more about the ideal gas here:

brainly.com/question/27691721

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3 0
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The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

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Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

5 0
3 years ago
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