Answer:
the question is incomplete
Answer:
0.0432M
Explanation:
We begin by writing a balanced equation for the reaction. This is illustrated below:
NaOH + HCl —> NaCl + H2O
From the equation above,
The number of mole of the acid (nA) = 1
The number of mole of the base (nB) = 1
Data obtained from the question include:
Vb (volume of the base) = 54mL
Cb (concentration of the base) = 0.1M
Va (volume of the acid) = 125mL
Ca ( concentration of the acid) =?
Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:
CaVa/CbVb = nA/nB
Ca x 125 / 0.1 x 54 = 1
Cross multiply to express in linear form:
Ca x 125 = 0.1 x 54
Divide both side by 125
Ca = (0.1 x 54) / 125
Ca = 0.0432M
Therefore, the concentration of the acid is 0.0432M
With the principle quantum number being 2, the maximum number that can share this is 8. You can use the general formula 2n^2 to calculate this number (n=quantum level), or you can use the concept of quantum numbers (n, l, m, s) to justify this answer.
The answer is
b) electron
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
