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dem82 [27]
3 years ago
12

When 1 mol of a fuel burns at constant pressure, it exchanges-3452 kJ of heat and does-11 kJ of workon the surroundings. What ar

e ΔUand ΔH for the combustion of thisfuel?
Chemistry
1 answer:
Degger [83]3 years ago
6 0

Answer:-3463 kJ and -3452kJ

Explanation:

ΔU is the change in internal energy of a system and its formula is;

ΔU = q + w

Where q represents heat transferred into or out of the system. Its value is positive when heat is transfer into the system and negative when heat is produced by the system.

W represents the work done on or by the system. Its value is positive when work is done on the system and negative when it is done by the system.

For the system in this question, we see that it produces heat which means heat is transferred out of the system, therefore the value of q is negative, it can also be seen that work is done by the system which means that w is also negative.

Therefore,

ΔU = -q-w

ΔU = -3452 kJ – 11kJ

= - 3463kJ

ΔH is the change in the enthalpy of a system and its formuls is;

ΔH = ΔU + Δ(PV)

By product rule Δ(PV) becomes ΔPV + PΔV

At constant pressure ΔP = 0. Therefore,

ΔH = -q-w + PΔV

w is equals to PΔV, So:

ΔH = -q

ΔH = -3452kJ

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Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.
Naddik [55]

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

  • molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M

3 0
3 years ago
How much heat in kj is released by burning 9.5 grams of methane?
tatyana61 [14]
Methane is the compound CH4, and burning it uses the reaction:

CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.

Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.

n = 9.5/16.042 = 0.592195 mol

Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
7 0
3 years ago
Label each carbon atom with the appropriate geometry. Bin 1 points to a carbon bonded to a double bonded carbon and single bonde
N76 [4]

Answer:

Bin 1 points to a carbon bonded to a double bonded carbon and single bonded to two hydrogens. --- trigonal planar, tetrahedral

Bin 2 points to a carbon double bonded to a carbon and single bonded to a carbon and one hydrogen.------- trigonal planar, tetrahedral

Bin 3 is a carbon single bonded to two carbons and single bonded to two hydrogens. ----- tetrahedral, tetrahedral

Bin 4 is the same as bin 3.--------tetrahedral, tetrahedral

Bin 5 is a carbon triple bonded to a carbon and single bonded to a carbon.---- linear, tetrahedral

Bin 6 is triple bonded to a carbon and single bonded to a hydrogen.---linear, tetrahedral

Explanation:

A single C-C or C-H bond is in a tetrahedral geometry, the carbon atom is bonded to four species with a bond angle of 109°.

A C=C bond is trigonal planar with a bond angle of 120°.

Lastly, a C≡C bond has a linear geometry with a bond angle of 180° between the atoms of the bond.

7 0
3 years ago
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0
2 years ago
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