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koban [17]
3 years ago
9

A Compound with Unknown Structure has

Chemistry
1 answer:
Nitella [24]3 years ago
7 0
Hi..........................
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One day on her walk home from school, Leann noticed some rust on an old car.
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Iron and Oxygen

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Fe is Iron and O2 is Oxygen

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Which of the following describes a condition in which an individual would not hear an echo?
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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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3 years ago
Which pH corresponds to the equivalence point for the graph below?
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Explication nose
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