<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
<u>Answer:</u> The empirical formula for the given compound is 
<u>Explanation:</u>
We are given:
Percentage of H = 5.80 %
Percentage of O = 23.02 %
Percentage of N = 20.16 %
Percentage of Cl = 51.02 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 5.80 g
Mass of O = 23.02 g
Mass of N = 20.16 g
Mass of Cl = 51.02 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Oxygen = 
Moles of Nitrogen = 
Moles of Chlorine = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.
For Hydrogen = 
For Oxygen = 
For Nitrogen = 
For Chlorine = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of H : O : N : Cl = 4 : 1 : 1 : 1
Hence, the empirical formula for the given compound is 
They both turn something on, and the way they are different is the way they turn it off<span>
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S2F6, is the correct answer I believe!
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:
A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4
As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.