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Alex787 [66]
3 years ago
5

Which of the following describes the sharing of a nonbonding electron pair on a nitrogen molecule with an oxygen atom, resulting

in a nitrous oxide molecule?
Chemistry
1 answer:
Gala2k [10]3 years ago
5 0
This is coordinate (dative) bonding where the nitrogen atom donates 2 electrons to the oxygen but is still chemically bonded.
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A 0.175 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.27° in a 1-dm sample
-BARSIC- [3]

Answer:

The specific rotation of D is 11.60° mL/g dm

Explanation:

Given that:

The path length (l) =  1 dm

Observed rotation (∝) = + 0.27°

Molarity = 0.175 M

Molar mass = 133.0 g/mol

Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol

Concentration in (g/mL) = 23.275 g/L

Since 1 L = 1000 mL

Concentration in (g/mL) = 0.023275 g/mL

The specific rotation [∝] = ∝/(1×c)

= 0.27°/( 1  dm ×  0.023275 g/mL )

= 11.60° mL/g dm

Thus, the specific rotation of D is 11.60° mL/g dm

3 0
3 years ago
From the dry desert of the south to the mountains of the north—Arizona has a varied topography. When residents of Phoenix visit
dedylja [7]

Answer:

The reason is because Flagstaff is at a higher elevation than Phoenix.

Explanation:

The air is thinner at higher elevations. You can google Flagstaff's elevation compared to Phoenix but the simple answer is that air is thinner at higher elevations and some people used to 'thicker' air find it harder to breath, especially after some strenuous exercise.

6 0
3 years ago
Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

5 0
3 years ago
Copper2 nitrate > nitrogen dioxide + copper2 oxide+ oxygen
erastovalidia [21]
Cu(NO3)2>NO2+CuO+O2 balanced: 2Cu(NO3)2=4NO2+2CuO+O2
3 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide (CO2)?
wariber [46]

Answer:

= 72.73%

Explanation:

The percentage by mass of an element is given by;

% element = total mass of element in compounds/molar mass of compound × 100

The mass of oxygen in carbon dioxide = 32 g

Molar mass of CO2 = 44 g

Therefore;

% of O2 = 32/44 × 100%

             <u>= 72.73%</u>

8 0
3 years ago
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