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Goryan [66]
3 years ago
13

In 1986 an electrical power plant in taylorsville, georgia, burned 8,376,726 tons of coal, a national record at that time. assum

ing that the coal was 80.7 % carbon by mass and that combustion was complete, calculate the number of tons of carbon dioxide produced by the plant during the year. if 71.0 % of the so2 could be removed by reaction with powdered calcium oxide, cao, via the reaction
Chemistry
2 answers:
RideAnS [48]3 years ago
6 0
<span>C + O2 → CO2 (8,376,726 tons) x (0.80) / (12.01078 g C/mol) x (1 mol CO2/ 1 mol C) x (44.00964 g CO2/mol) = 24,555,054 tons CO2</span>
stellarik [79]3 years ago
5 0

Answer:

Carbon dioxide is produced, and as a gas, it was not captured to measure its mass.

Explanation:

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent
elixir [45]

Answer : The initial rate for a reaction will be 3.8\times 10^{-4}Ms^{-1}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

A+B+C\rightarrow P

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1

k=7.6\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1

\text{Rate}=3.8\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.8\times 10^{-3}Ms^{-1}

6 0
3 years ago
Which part of the experiment is not touched by the independent variable or is the normal/comparison
kobusy [5.1K]
The part of the experiment that’s is not touched by the independent variable and is for comparison is called the :
Control Group
8 0
3 years ago
PLEASE HELP!!!!!!!
Keith_Richards [23]

Answer: The gas generated by two antacid tablets has a smaller volume.

Explanation:

Since the antiacid is the limiting reagent, we know that the more tablets there are, the more gas there will be.

This means that there will be more gas generated by the four antiacid tablets when compared to the two antiacid tablets, which gives us that the gas generated by the two antiacid tablets has a smaller volume.

6 0
2 years ago
When the pH of a solution is 12.83, what is [H +]?
inn [45]

Answer:

B.9.710-11M

Explanation:

<h3>plss tell me if im wrong</h3>

4 0
3 years ago
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts
Y_Kistochka [10]

Answer:

Explanation:

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?

Mg + O2 → MgO (unbalanced)

first, balance the equation

2Mg +O2-------> 2MgO

two magnesium atoms react with one diatomic oxygen molecule

there is a 1:1 ratio of magnesium to oxygen atoms

but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms

the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.

6 0
2 years ago
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