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antoniya [11.8K]

3 years ago

12

The process of manufacturing sulfuric acid begins with the burning of sulfur. What mass of sulfur would have to be burned in ord

er to produce 1.00 kg of H2SO4 ? Assume that all of the sulfur ends up in the sulfuric acid.

1 answer:

Minchanka [31]3 years ago

5 0

Answer:

about 327 g of sulfur

Explanation:

Reaction equations.

2S + 2O2 = 2SO2

2SO2 + O2 = 2SO3

2SO3 + 2H2O = 2H2SO4

1.00 kg H2SO4 (1000 g H2SO4 / 1kg H2SO4) x (1 mole H2SO4 / 98.079 g H2SO4) x

(1 mole S / 1 mole H2SO4) x (32.065 g S / 1 mole S)

= 326.93 g S needed to be burned to produce 1.00 kg of H2SO4

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A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as

umka2103 [35]

Answer:

I would expect to extract the acetic acid.

Explanation:

In the first step, since we are adding a concentrated acid,<u> it will react with the bases present in the mixture (diethylamine and ammonia) </u><u>forming salts</u><u>, </u><u>which are soluble in water</u>. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.

In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.

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3 years ago

For the reaction PC15 (8) PC13 (g) + Cl2 (g) K = 0.0454 at 261 °C. If a vessel is filled with these gases such that the initial

Fiesta28 [93]

Answer:

The correct answer is B.

The $K_{eq}$ is samller than $Q$ of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression for $Q$ is written as:

$Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}$

$Q=\frac{0.20 M\times 2.5 M}{0.20 M}$

$Q=2.5$

Given : $K_{eq}$ = 0.0454

Thus as $K_{eq}$ , the reaction will shift towards the left i.e. towards the reactant side.

4 0

3 years ago

Read 2 more answers

How do we measure the mass of an extrasolar planet?

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4 0

3 years ago

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [

sveta [45]

Explanation:

The given data is as follows.

$E^{o}$ = 0.483, $[Co^{3+}]$ = 0.173 M,

$[Co^{2+}]$ = 0.433 M, $[Cl^{-}]$ = 0.306 M,

$P_{Cl_{2}}$ = 9.0 atm

According to the ideal gas equation, PV = nRT

or, P = $\frac{n}{V}RT$

Also, we know that

Density = $\frac{mass}{volume}$

So, P = MRT

and, M = $\frac{P}{RT}$

= $\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}$

= $\frac{9.0}{24.436}$

= 0.368 mol/L

Now, we will calculate the cell potential as follows.

E = $E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}$

= $0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}$

= $0.483 - 0.02955 log \frac{0.0689}{0.0162}$

= $0.483 - 0.02955 \times 0.628$

= 0.483 - 0.0185

= 0.4645 V

Thus, we can conclude that the cell potential of given cell at $25^{o}C$ is 0.4645 V.

4 0

3 years ago