add up the mass of protons and neutrons
Answer:
Halocarbon, any chemical compound of the element carbon and one or more of the halogens (bromine, chlorine, fluorine, iodine); two important subclasses of halocarbons are the chlorocarbons, containing only carbon and chlorine, and the fluorocarbons, containing only carbon and fluorine.
Explanation:
Answer:
3.40g/mL
Explanation:
Density is a measure of mass over volume, so to get the density all we have to do is divide the mass by the volume.
21.35g ÷ 6.28 mL = 3.40g/mL
Answer:
The answer to your question is pH = 0.686
Explanation:
Data
Acetic acid = 20 ml
Final volume = 1.7 L
pH = ?
density = 1.05 g/ml
Process
1.- Calculate the mass of acetic acid
density = mass / volume
mass = density x volume
mass = 1.05 x 20
mass = 21 g
2.- Calculate the moles of acetic acid (CH₃COOH)
Molar mass = (12 x 2) + (16 x 2) + (4 x 1)
= 24 + 32 + 4
= 60 g
60 g of acetic acid ---------------- 1 mol
21 g of acetic acid ----------------- x
x = (21 x 1) / 60
x = 0.35 moles
3.- Calculate the concentration
Molarity = 0.35 moles / 1.70 L
= 0.206
4.- Calculate the pH
pH = -log[0.206]
pH = 0.686
The wt% of KOH = 45%
This implies that there is 45 g of KOH in 100 g of the solution
Density of the solution is given as 1.45 g/ml
Therefore, the volume corresponding to 100 g of the solution is
= 100 g * 1 ml /1.45 g = 68.97 ml = 0.069 L
Now concentration of the concentrated KOH solution is:
Molarity = moles of KOH/vol of solution
= (45 g/56.105 g.mol-1)/0.069 L = 11.6 M
Thus,
Initial KOH concentration M1 = 11.6 M
Initial volume = V1
Final concentration M2 = 1.20 M
Final volume V2 = 250 ml
M1*V1= M2*V2
V1 = M2*V2/M1 = 1.20*250/11.6 = 25.9 ml = 26 ml
