The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
Answer:
Reactions 1, 3 and 5
Explanation:
First thing's first, let's ensure that all the reactions given are balanced. This is given as;
CO(g) + 1/2 O2(g )→ CO2(g)
Li(s) + 1/2 F2(l) → LiF(s)
C(s) + O2(g) → CO2(g)
CaCO3(g) → CaO + CO2(g)
2Li(s) + F2(g) → 2LiF(s)
For the condition to be valid;
- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.
- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.
This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.
Answer:
Explanation:
Hello,
In this case, considering that the by-mass percent of water is:
Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:
Best regards.
Given two electrons with charge of 1.5x10^-10 m
The electostatic force between them is determined by this formula:
F = kq1q2/r^2
where
k = 9x10^9
q1 = q2 = 1.5x10^-10
r = 2.82x10^-15
F = 9x10^9 * (1.5x10^-10)^2 / 2.82x10^-15
= 71808.51
Answer:
The pH of the solution is 8.
Explanation:
To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:
1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:
pOH = –Log [OH-]
pOH = 6
6 = –Log [OH-]
–6 = Log [OH-]
[OH-] = Antilog (–6)
[OH-] = 1x10^–6 mol/L
2. The pH of the solution can be obtained as follow:
pH + pOH = 14
pOH = 6
pH + 6 = 14
pH = 14 – 6
pH = 8.
From the calculations made above,
[OH-] = 1x10^–6 mol/L
pH = 8.
Therefore, the correct answer is:
The pH of the solution is 8
