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mrs_skeptik [129]

3 years ago

12

The enthalpy of combustion of gasoline (C8H18) is 5471 kJ mol-1. How much heat will be produced by burning 1 gallon of gasoline

at 25°C? (Density of gasoline = 0.703 g/mL; 1gallon = 3.78 L) Show your work clearly.

1 answer:

kvv77 [185]3 years ago

8 0

Answer:

127529 KJ

Explanation:

Since 1gallon = 3.78 L= 3780 ml

The density of C8H18= 0.703 g/mL

Density = mass/volume

Mass= Density × volume

Mass= 0.703 g/mL × 3780 ml

Mass= 2657.34 g

Molar mass of C8H18= 114 g/mol

Number of moles= mass/molar mass

Number of moles= 2657.34/114

Number of moles= 23.31 moles

Since 1 mole evolved=5471 kJ

23.31 moles = 5471 kJ × 23.31 = 127529 KJ

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Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

$4.67 \times 10^{-6} = x \times (0.10)^{2}$

x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

5 0

3 years ago

Read 2 more answers

Let's say that you were given the following

mezya [45]

Answer:

V2~0.4839M

Explanation:

We're going to use Boyles law to answer the question.

Boyle's law:

P1V1=P2V2

P1=151mmHg

P2=166mmHg

V1=0.532L

V2=?

V2=(P1 x V1)/P2

V2=(151 x 0.532)/166

V2~0.4839M

Hope it helps:)

8 0

3 years ago

What is an element considered a pure substance?

Alona [7]

A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.

8 0

3 years ago

Chemistry. Will mark Brainly.

kogti [31]

Answer:

c is right

Explanation:

8 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago