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Chemistry

1 answer:

IgorLugansk [536]3 years ago

6 0

Answer:

Explanation:

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Explain why bond angle of H2O is greater than OF2

atroni [7]

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.

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A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

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CVOPEDIA-CVOSOFT

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Seans formula did not have any atmos of oxygen

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What are you asking???? If the formula had no atoms of oxygen then......

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Do step 3 as outlined in the lab guide. Record your results in the appropriate blanks.

PolarNik [594]

Explanation:

Do the step 3 as outlined in the lab guide. record your results in the appropriate blank.

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