Answer:
22:
Formular:

substitute:

23:
<em>Same</em><em> </em><em>element</em><em> </em><em>is</em><em> </em><em>represented</em><em> </em><em>by</em><em> </em><em>same</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>protons</em><em>.</em><em> </em>
Answer:
6 protons. 6 protons
7 neutrons. 8 neutrons
6 electrons. 6 electrons
Note: <u>Atoms</u><u> </u><u>with</u><u> </u><u>same</u><u> </u><u>proton</u><u> </u><u>number</u><u> </u><u>but</u><u> </u><u>different</u><u> </u><u>mass</u><u> </u><u>number</u><u> </u><u>are</u><u> </u><u>called</u><u> </u><u>isotopes</u>
1. Meteorologist predict the weather by using tools. They use these tools to measure atmospheric conditions that occurred in the past and present, and they apply this information to create educated guesses about the future weather. The best we can do is observe past and present atmospheric patterns and data, and apply this information to what we think will happen in the future. Meteorologists use the scientific method on a daily – and even hourly – basis!
2. They use thermometers, barometers, sling psychrometers and rain gauges. They also use anemometers, hygrometers, weather maps, weather balloons and weather satellites.
Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
This reaction is most likely to fall under SN2 because the
thing called carbonication does not occur in SN1. The carbon forms a partial
bond with the nucleophile during the intermediate phase and the leaving group.
So for this question the reaction will fall under SN2.