Atomic number is your answer.
Answer:
Choice B, C, and D.
Explanation:
Choice A is not true in general. Here's a way to think about that. Consider a very special equilibrium where the concentration of reactants and products are indeed equal. When one of the external factors (such as temperature) changes, the equilibrium will shift towards either side of the reaction. More products will be converted to reactants, or vice versa. Either way, in the new equilibrium, the concentration of the reactants and products will not be equal any more.
Choice B should be considered with choice C and D in mind.
Choice C is indeed correct. The reaction rate would not be zero unless all the reactants were used up or taken out of the system. That's not what happens in an equilibrium. Instead, when reaction rate is plotted against time, the graph for reactions in both directions will eventually flat out at a non-zero value.
Choice D explains why even though choice C is correct, the concentration of a system at equilibrium stays the same. At the equilibrium, reactions in both directions are still happening. However, during the time it takes for the forward reaction use up some reactant particles, the reverse reaction would have produced these particles again. On a large scale, there would be no observable change to the concentration of each species in the equilibrium. Therefore, choice B is also correct.
Answer: (not sure)
T - A
T - A
T - A
A - T
C - G
G - C
C - G
C - G
A - T
im not sure about the amoni acid produced but
AAA produces -> Lys.
ACG produces -> Thr.
and i don't know about the last one
1.9993999057621666 moles
1 mole = 36.46094
Answer:
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Explanation:
Step 1: Data given
The solution contains 0.021 M Cl- and 0.017 M I-.
Ksp(CuCl) = 1.0 × 10-6
Ksp(CuI) = 5.1 × 10-12.
Step 2: Calculate [Cu+]
Ksp(CuCl) = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [0.021]
[Cu+] = 1.0 * 10^-6 / 0.021
[Cu+] = 4.76 *10^-5 M
Ksp(CuI) = [Cu] [I]
5.1 * 10^-12 = [Cu+] [I-]
5.1 * 10^-12 =[Cu+] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017
[Cu+] = 3.0 *10^-10 M
[Cu+]from CuI hast the lowest concentration
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
