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Liula [17]
2 years ago
6

What is the total number of pairs of electrons that one carbon atom shares?

Chemistry
1 answer:
wel2 years ago
8 0

Answer:

4

Explanation:

A carbon atom has 4 electrons in its outermost shell (2s^2p^2).  All are unpaired (none share their orbital with another electron).  So all four are anxious to pair with another electron.  Once it has found 4 more electrons contributed from other atom(s), it will have 4 pairs of shared electrons.

Hydrogen has one lone electron.  An atom of H is downright gleeful in sharing it's electron with elements such as carbon, C.  Since carbon has 4 unpaied electrons, it will combine with 4 H atoms.  At that point, cabon is sharing 4 electron pairs.

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10. A compound has an empirical formula of CH, and a molecular mass of 6 points
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Answer:

Molecular formula = C₅H₁₀

Explanation:

From the question given above, the following data were obtained:

Empirical formula of compound => CH₂

Molar mass of compound = 70.1 amu

Molecular formula of compound =...?

The molecular formula of a compound is usually a multiple (n) of the empirical formula i.e

Molecular formula = [CH₂]ₙ

Thus, to obtain the molecular formula of the compound, we must first determine the value of n. The value of n can be obtained as follow:

[CH₂]ₙ = 70.1

[12 + (2×1)]n = 70.1

[12 + 2]n = 70.1

14n = 70.1

Divide both side by 14

n = 70.1 / 14

n = 5

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₅

Molecular formula = C₅H₁₀

Thus, the Molecular formula of the compound is C₅H₁₀.

Explanation:

HOPE THIS HELPS!

7 0
3 years ago
A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
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<u>Answer:</u> The concentration of KOH solution is 1.215 M

<u>Explanation:</u>

For the given chemical equation:

2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50M\\V_1=16.2mL\\n_2=1\\M_2=?M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 1.50\times 16.2=1\times M_2\times 40.0\\\\M_2=\frac{2\times 1.50\times 16.2}{1\times 40.00}=1.215M

Hence, the concentration of KOH solution is 1.215 M

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