When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
Answer: Molarity of 
anions in the chemist's solution is 0.0104 M
Explanation:
Molarity : It is defined as the number of moles of solute present per liter of the solution.
Formula used :
where,
n= moles of solute
= volume of solution in ml = 100 ml
Now put all the given values in the formula of molarity, we get
Therefore, the molarity of solution will be 
As 1 mole of 
gives 2 moles of
Thus 
moles of gives =
Thus the molarity of anions in the chemist's solution is 0.0104 M
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Correct Answer: compound because the atoms of the elements are combined in a fixed proportion.
