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natali 33 [55]
1 year ago
8

For each pair of bonds, indicate the more polar bond, and use an arrow to show the direction of polarity in each bond.a. C-O and

C-N
Chemistry
1 answer:
Naddika [18.5K]1 year ago
7 0

Answer:

The more polar bond is C-O.

Explanation:

The greater polarity is due to that the Oxygen atom is more electronegative than the Nitrogen atom, so the negativity of the molecule tends to be on that side.

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Avagadros Law..<br>Definition please..​
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Answer:

"Avogadro's law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The law is a specific case of the ideal gas law. A modern statement is: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

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3 years ago
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Question 5(Multiple Choice Worth 3 points)
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Answer:

\sqrt{ | {o2}^{2} | }  \times \frac{?}{?}  \sqrt[?]{?}  \times \frac{?}{?}  \sqrt[ |?| ]{?}  \sqrt{?}  \times  \frac{?}{?}

Explanation:

exactly

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3 years ago
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

8 0
3 years ago
The percent of remaining parent isotope in a radioactive decay process is 40 percent. How many half-lives have elapsed since the
muminat

Answer: Between 1 and 2.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

a=\frac{a_o}{2^n}        ............(1)

where,

a = amount of reactant left after n-half lives  = 40

a_o = Initial amount of the reactant  = 100

n = number of half lives

Putting in the values we get:

40=\frac{100}{2^n}  

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taking log on both sides

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Thus half-lives that have elapsed is between 1 and 2

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