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3 years ago

Explain why bond angle of H2O is greater than OF2

1 answer:

7 0

It's only a small difference (103 degrees versus 104 degrees in water), and I believe the usual rationalization is that since F is more electronegative than H, the electrons in the O-F bond spend more time away from the O (and close to the F) than the electrons in the O-H bond. That shifts the effective center of the repulsive force between the bonding pairs away from the O, and hence away from each other. So the repulsion between the bonding pairs is slightly less, while the repulsion between the lone pairs on the O is the same -- the result is the angle between the bonds is a little less.

Hope this helps!

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klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
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$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
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3 years ago