Based on the liquids viscosity (which is a measure of how quickly/slowly a fluid can flow).
Higher viscosity represents a "thicker" liquid & slower flow (honey)
Lower viscosity represents a "thinner" liquid & quicker flow (vinegar)
Answer:
- The abundance of 107Ag is 51.5%.
- The abundance of 109Ag is 48.5%.
Explanation:
The <em>average atomic mass</em> of silver can be expressed as:
107.87 = 106.90 * A1 + 108.90 * A2
Where A1 is the abundance of 107Ag and A2 of 109Ag.
Assuming those two isotopes are the only one stables, we can use the equation:
A1 + A2 = 1.0
So now we have a system of two equations with two unknowns, and what's left is algebra.
First we<u> use the second equation to express A1 in terms of A2</u>:
A1 = 1.0 - A2
We <u>replace A1 in the first equation</u>:
107.87 = 106.90 * A1 + 108.90 * A2
107.87 = 106.90 * (1.0-A2) + 108.90 * A2
107.87 = 106.90 - 106.90*A2 + 108.90*A2
107.87 = 106.90 + 2*A2
2*A2 = 0.97
A2 = 0.485
So the abundance of 109Ag is (0.485*100%) 48.5%.
We <u>use the value of A2 to calculate A1 in the second equation</u>:
A1 + A2 = 1.0
A1 + 0.485 = 1.0
A1 = 0.515
So the abundance of 107Ag is 51.5%.
Answer:
Iodide> Bromide > chloride > flouride
Explanation:
During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.
This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).
In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.
Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.
Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride
By space travel.
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