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Solnce55 [7]
3 years ago
5

Carbon disulfide is an important industrial solvent. It is prepared by the reaction of carbon (called coke) with sulfur dioxide.

Carbon monoxide is also produced (as a byproduct) of this reaction. (a) Write the equation that correctly demonstrates this reaction and balance it. (b) How many moles of carbon are needed to produce 45.90 moles of carbon disulfide?
Chemistry
1 answer:
balu736 [363]3 years ago
6 0

Answer:

A. The balanced equation is given below:

5C + 2SO2 —> CS2 + 4CO

B. 229.5 moles of carbon.

Explanation:

A. Step 1:

The equation for the reaction is given below:

C + SO2 —> CS2 + CO

A. Step 2:

Balancing the equation.

The above equation can be balanced as follow:

There are 2 atoms of S on the right side and 1 atom on the left side. It can be balance by putting 2 in front of SO2 as shown below:

C + 2SO2 —> CS2 + CO

There are 4 atoms of O on the left side and 1 on the right. It can be balance by putting 4 in front of CO as shown below:

C + 2SO2 —> CS2 + 4CO

There are a total of 5 atoms of C on the right side and 1 atom on the left. It can be balance by putting 5 in front of C as shown below:

5C + 2SO2 —> CS2 + 4CO

Now the equation is balanced

B. Determination of the number of moles of carbon needed to produce 45.90 moles of carbon disulfide.

This is illustrated below:

5C + 2SO2 —> CS2 + 4CO

From the balanced equation above,

5 moles of C produced 1 mole of CS2.

Therefore, Xmol of C will produce 45.90 moles of CS2 i.e

Xmol of C = 5 x 45.90

Xmol of C = 229.5 moles

229.5 moles of carbon is needed to produce 45.90 moles of carbon disulfide.

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How many significant figures<br> are in this number?<br> 43.55
leva [86]

Answer:

4 significant figures

Explanation:

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All digits (except for 0) are always significant. Therefore, all the digits in 43.55 are significant. Since there are 4 digits in the given number, there are 4 significant figures.

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1 year ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
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Answer:

0.382g

Explanation:

Step 1: Write the reduction half-reaction

Al³⁺(aq) + 3 e⁻ ⇒ Al(s)

Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s

We will use the following relationships.

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The mass of Al produced is:

41.0s \times \frac{100C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molAl}{3mole^{-} } \times \frac{26.98gAl}{1molAl} = 0.382gAl

7 0
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Balance each of these equations.
mafiozo [28]
The answer I would choose is the third one
4 0
3 years ago
(04.01 LC)
Marysya12 [62]

Answer:

Substance at the beginning of a reaction- reactant

Substance at the end of a reaction- product

Number placed before a compound in a chemical equation- stoichiometric coefficient

Explanation:

In a reaction equation, the species written on the left hand side of the equation are called the reactants.

The reactants combine to form the species on the right hand side of the reaction equation called products.

The stoichiometric coefficient is a number written before the formula of a compound in the reaction equation.

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