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KATRIN_1 [288]
3 years ago
7

Convert 5.00 moles of carbon to grams. Be sure to report your answer with the correct significant figures.

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
8 0

1 moles Carbon to grams = 12.0107 grams

2 moles Carbon to grams = 24.0214 grams

3 moles Carbon to grams = 36.0321 grams

4 moles Carbon to grams = 48.0428 grams

5 moles Carbon to grams = 60.0535 grams

6 moles Carbon to grams = 72.0642 grams

7 moles Carbon to grams = 84.0749 grams

8 moles Carbon to grams = 96.0856 grams

9 moles Carbon to grams = 108.0963 grams

10 moles Carbon to grams = 120.107 grams

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Read 2 more answers
A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
3 years ago
You need a 35% alcohol solution. On hand, you have a 280 mL of a 30% alcohol mixture. You also have 75% alcohol mixture. How muc
Marina CMI [18]

Answer:

35 mL

Explanation:

Let the amount of 75% mixture needed be A.

The amount of each solution with their respective concentration to be added together can be expressed as:

0.75A + 0.3(280)...............eqn 1

The addition of the two solution must merge with the expected concentration and this can be expressed as:

0.35(A + 280).............eqn 2

Eqn 1 must be equal to eqn 2, hence:

0.75A + 0.3(280) = 0.35(A + 280)

Solve for A.

0.75A + 84 = 0.35A + 98

0.40A = 14

A = 35

Hence, 35 mL of the 75% mixture will be needed.

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3 years ago
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Answer:

d

Explanation:

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