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2 years ago

A benzene bonded side group is called?

Chemistry

1 answer:

hammer [34]2 years ago

3 0

It is called a phenyl group

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A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

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Homogeneous means composed of parts of different kinds; having widely different elements or components.

Explanation:

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