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Tju [1.3M]
3 years ago
12

Properties of carbon responsible for the pressence of millions of organic compound ​

Chemistry
1 answer:
Minchanka [31]3 years ago
5 0

Answer:

Carbon has the ability to form very long chains of interconnecting C-C bonds. This property allows carbon to form the backbone of organic compounds, carbon-containing compounds, which are the basis of all known organic life. Nearly 10 million carbon-containing organic compounds are known.

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Give me lil reasoning so I know your not lying for points
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Answer:

0.01 psi

Explanation:

If you look at the data points plotted on the graph, the slope of the line touches 0.1 for the y-axis when it is at 20 for the x-axis.

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2 years ago
Given the balanced equation representing a
zhannawk [14.2K]

Answer: it would be a 1 to 1 ratio

Explanation: originally it would be 2 to 2 but you have to reduce

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Blood found in the circulatory system transports many materials, including nutrients, gases and wastes. Which of the following b
marysya [2.9K]

Answer: The Excretory system

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3 years ago
If I need 2.2 moles of CO2 , and I have excess Fe2O3 , how many moles of C do I need?
olchik [2.2K]

Answer:

0.733 mol.

Explanation:

  • From the balanced equation:

<em>2Fe₂O₃ + C → Fe + 3CO₂,</em>

It is clear that 1.0 moles of Fe₂O₃  react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.

  • Since Fe₂O₃ is in excess, C will be the limiting reactant.

<u><em>Using cross multiplication:</em></u>

1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.

??? mole of C produces → 2.2 moles of CO₂.

∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.

6 0
2 years ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
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