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3 years ago

Which one goes first

Chemistry

2 answers:

bezimeni [28]3 years ago

5 0

I think it's producers. Not sure, but I would think that something would have to be made before changes can take place.

NNADVOKAT [17]3 years ago

3 0

Producers then consumers then decomposers is the correct order from first to last.

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Please help help I need help ASAP What can you infer about the heartbeat when the pulse rate increases?

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Answer:

The heart is pumping more blood throughout your body

Explanation:

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3 years ago

The most common element in the atmosphere are

Ivanshal [37]

Nitrogen and oxygen are the most prevalent in the atmosphere.

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4 years ago

Dichlorine heptoxide (C1 2 0 7) is a

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Answer:

Identify the "given" information and what the problem is asking you to "find."

Given : Cl2O7

Find: % Composition (% Cl and %O)

List other known quantities.

Mass of Cl in 1 mol Cl2O7 , 2 Cl : 2 x 35.45 g = 70.90 g

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3 years ago

4) The most electronegative atom of the ones listed below is ____.AlBTlInGa

marusya05 [52]

Answer:

$B$

Explanation:

Here, we wnat to get the most electronegative atom

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The elements listed are in the same group

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Boron is the most electronegative atom in the group

5 0

1 year ago

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;

(a) Balanced equation including N_2 from air

The balanced equation ignoring 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

Including N_2 from air, the balanced equation is

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

(b) Balanced equation for 120 % stoichiometric combustion

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

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) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

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