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aliina [53]
4 years ago
14

explain why the hydrogen atoms in a hydrogen gas molecule form nonpolar covalent bonds but oxygen and hydrogen atoms in water mo

lecules form polar covalent bonds
Chemistry
1 answer:
marishachu [46]4 years ago
7 0

Answer:

Explanation:

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards itself and becomes partial negative while the other atom becomes partial positive.

Non polar covalent bond:

It is the bond where both bonded atoms share the pair of electron equally.

For example:

Hydrogen gas (H₂) is non polar covalent compound because the electronegativity of both bonded atoms are same. No poles are created that's why this is non polar covalent compound.

Polar covalent bond:

It is the bond where both bonded atoms share the pair of electron unequally.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive and bond is polar.

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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

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4 years ago
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Answer:

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Explanation:

H+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

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