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2 years ago

10. Which is an example of mechanical energy?*

Chemistry

1 answer:

gtnhenbr [62]2 years ago

6 0

Answer: A golfer hitting a golf ball.

Explanation:

The atomic particles move more in this option than the others.

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Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

3 0

2 years ago

A welding torch requires 602.1 L of ethylene gas at 2.77 atm. What will be the pressure of the gas if ethylene is supplied by a

Katarina [22]

PV=PV
(602.1 L)(2.77atm) = (110.6 L) (X atm)
1667.817=110.6X
15.07971971 atm = X
Rounds to 15.1 (sig figs so much fun)

8 0

1 year ago

Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2

Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

ΔG°rxn = ΔH°rxn - T×S°rxn (1)

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

ΔG°rxn = +50.8 kJ/mol

7 0

2 years ago

Read 2 more answers

Which element(s) are not balanced in this equation ?

Alekssandra [29.7K]

Only the Fe is unbalanced.
:)

3 0

2 years ago

Read 2 more answers

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

sergeinik [125]

The formula for molality---> m = moles solute/ Kg of solvent

the solute here is NH₃ because it's the one with less amount. which makes water the solvent.

1) let's convert the grams of NH₃ to moles using the molar mass

molar mass of NH₃= 14.0 + (3 x 1.01)= 17.03 g/ mol

15.0 g (1 mol/ 17.03 g)= 0.881 mol NH₃

2) let's convert the grams of water into kilograms (just divide by 1000)

250.0 g= 0.2500 kg

3) let's plug in the values into the molality formula

molality= mol/ Kg---> 0.881 mol/ 0.2500 kg= 3.52 m

6 0

2 years ago