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Shalnov [3]
3 years ago
13

If you need to reverse the following reaction and multiply it by 2 in order for it to be an intermediate reaction in a Hess's la

w problem, what would be the final value for the enthalpy of reaction you use for this intermediate reaction? H2 + 0.5 O2 H2O, H = -286 kJ
Chemistry
1 answer:
jekas [21]3 years ago
4 0
We have that the total enthalpy of the reaction changes with the quantity of the reactants and it is proportional to them. Also, the reverse of a reaction has the opposite enthalpy. Hence, since we need to multiply by 2, the reactants are double and thus the value of the enthalpy is 2 as big. Also, since we are using the inverse reaction, we must also invert the sign. Thus, for this reaction we must use the value H=572 kJ.
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irinina [24]

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Certain crops require a specific amount of annual rainfall to be successful. Which statement explains why coconut palm trees are
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Read 2 more answers
What is the diameter of Neptune in standard notation
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Answer:

The diameter of Neptune is approximately 49,500 km. This makes Neptune the 4th largest planet in the Solar System, after Jupiter, Saturn and Uranus.

7 0
3 years ago
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder
IRISSAK [1]

Answer:

A. Up

B. Out

C. Out

D. To equilibrum

Explanation:

a. The reaction in an exothermic reaction so this means heat is given off. If the cylinder is thin enough heat will transfer to the water bath

b. Since the products will create heat which will increase pressure, the piston in an attempt to maintaining a constant pressure will move up to accommodate building pressure.

c. Heat will flow out of the gaseous mixture as this reaction creates heat as a product as well

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7 0
3 years ago
At 373.15K and 1 atm, the molar volume of liquid water and steamare 1.88 X 10-5 m3 and 3.06 X 10-2m3, respectively. Given that t
professor190 [17]

Answer The value of \Delta H and \Delta U is, 40.79 kJ and 37.7 kJ respectively.

Explanation :

Heat released at constant pressure is known as enthalpy.

The formula used for change in enthalpy of the gas is:

\Delta Q_p=\Delta H=40.79kJ/mol

Now we have to calculate the work done.

Formula used :

w=-P\Delta V\\\\w=-P\times (V_2-V_1)

where,

w = work done  = ?

P = external pressure of the gas = 1 atm

V_1 = initial volume = 1.88\times 10^{-5}m^3=1.88\times 10^{-5}\times 10^3L=1.88\times 10^{-2}L

V_2 = final volume = 3.06\times 10^{-2}m^3=3.06\times 10^{-2}\times 10^3L=3.06\times 10^{1}L

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

w=-(1atm)\times (3.06\times 10^{1}-1.88\times 10^{-2})L

w=-30.5812L.atm=-30.5812\times 101.3J=-3097.87556J=-3.09\times 10^3J=-3.09kJ

Now we have to calculate the change in internal energy.

\Delta U=q+w

\Delta U=40.79kJ+(3.09kJ)

\Delta U=37.7kJ

Thus, the value of \Delta H and \Delta U is, 40.79 kJ and 37.7 kJ respectively.

4 0
3 years ago
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