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ANTONII [103]
3 years ago
8

A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will

be reacted with calcium hydroxide (slaked lime), which costs $0.25 per pound. Write the balanced chemical equation for this process. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) Determine the cost that the firm will incur from this use of slaked lime. Cost
Chemistry
1 answer:
Reil [10]3 years ago
3 0

Answer:

The balanced chemical equation for this process:

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

$194.51 is the cost that the firm will incur from this use of slaked lime.

Explanation:

The balanced chemical equation for this process

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

Moles of sulfuric acid = n

Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L

1 gallon = 3.785 Liter

Morality of the sulfuric acid = M = 1.05 m

Molarity(M)=\frac{n}{V(L)}

n=m\times V=1.05 M\times 4,542 L=4,769.1 mol

According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;

\frac{1}{1}\times 4,769.1 mol=4,769.1 mol of calcium hydroxide

Mass of 4,769.1 moles of calcium hydroxide:

4,769.1 mol  74 g/mol = 352,913.4 g

= \frac{352,913.4 }{453.6} pounds =778.03 pounds

(1 pound = 453.6 grams)

Cost of 1 pound of slaked lime  = $0.25

Cost of 778.03 pounds of slaked lime  = $0.25 × 778.03 = $194.51

$194.51 is the cost that the firm will incur from this use of slaked lime.

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Solution : Given,

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3 years ago
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Write the name for the following molecular compounds. Remember to use the correct prefix for each compound.
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3 0
3 years ago
Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.
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Answer:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

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E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt  = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

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5 0
3 years ago
How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of crcl3 using a current of 1.5 a?
PSYCHO15rus [73]
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        n = (1 g Cr)(1 mol Cr/51.996 g Cr)
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Determine the number in charge by multiplying with Faraday's constant,

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Then, calculate time by dividing the charge with the current,

     t = 5566.87 C/1.5 A 
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     t = 61.84 hours

<span><em>Answer: 61.84 hours</em></span>


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