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bagirrra123 [75]
3 years ago
7

Write the electron configuration for Silicon

Chemistry
1 answer:
morpeh [17]3 years ago
5 0

Answer:

Si₁₄ = 1s² 2s² 2p⁶ 3s² 3p²

Explanation:

Silicon is present in group 14 of periodic table.

It is blue-gray color metalloid.

Its atomic mass is 28 g/mol.

It is mostly used to make allow.

The most important alloy are Al-Si and Fe-Si. These are used to make different machine tools, transformer plates, engine etc.

Its atomic number is 14 and electronic configuration can be written as,

Electronic configuration:

Si₁₄ = 1s² 2s² 2p⁶ 3s² 3p²

The noble gas electronic configuration or abbreviated electronic configuration can also be written.

Si₁₄ = [Ne] 3s² 3p²

The atomic number of neon is 10. Its electronic configuration is,

Ne₁₀ = 1s² 2s² 2p⁶

That's why we write [Ne] for 1s² 2s² 2p⁶ in abbreviated electronic configuration of Si.

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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
Drag and drop the gases in the atmosphere to order them from most abundant to least abundant.
Alexeev081 [22]
So what is the question
4 0
3 years ago
Read 2 more answers
How many moles of calcium atoms are in each mole of calcium carbonate?
Tanya [424]

Answer:

The number of moles of calcium atom in each mole of calcium carbonate is 1 mole of calcium atoms

Explanation:

The chemical formula for CaCO₃ = 100.086 g

The number of molecules in one mole of a substance is given by the Avogadro's number, N_A = 6.022 × 10²³ molecules

The number of calcium atoms in one molecule of CaCO₃ = 1 atom of calcium

Therefore, the number of calcium atoms in 1 mole of CaCO₃ = 1 × 6.022 × 10²³ atoms of calcium = 6.022 × 10²³ atoms of calcium

6.022 × 10²³ atoms of calcium = The number of toms in 1 mole of calcium atoms

Therefore, the number of moles of calcium atom in each mole of calcium carbonate = 1 mole of calcium atoms.

5 0
3 years ago
um bequer muito utilizado em laboratorio mede volumes de ate 250 ml.determine esse volume em: a)centimetros cubicos b)litros c)m
Lelechka [254]
Can u put it in english not be racist but i love chemistry i just dont know foreign language
6 0
4 years ago
Which best describes nitrogen fixation?
Mashcka [7]

Answer:

Nitrogen fixation is the process of converting nitrogen to a usable form.

5 0
3 years ago
Read 2 more answers
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