Answer:
v = 534.5mL
m = 597.15g
Density = 9.23g/mL
Density = 9.125g/mL
Explanation:
Density = mass/ volume
For the first question
Density = 1.59g/mL
Mass = 834.01g
Volume = ?
Using the above formula we have 1.59 = 834.01/v
v = 834.01/1.59
v = 534.5mL
For the second question
Density =0.9167g/mL
Volume = 651.41mL
Mass =?
Using the above formula we have
0.9167 =m/651.41
Cross multiply
m = 0.9167 x 651.41
m = 597.15g
For the third question
Mass =803.44g
Volume=87.03mL
Density =?
Density = 803.44/87.03
= 9.23g/mL
For the fourth
Density = 56.85/6.23
= 9.125g/mL
Answer:
The concentration of this sodiumhydroxide solutions is 0.50 M
Explanation:
Step 1: Data given
Mass of sodium hydroxide (NaOh) = 8.0 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Volume water = 400 mL = 0.400 L
Step 2: Calculate moles NaOH
Moles NaOH = mass NaOH / molar mass NaOH
Moles NaOH = 8.0 grams / 40.0 g/mol
Moles NaOh = 0.20 moles
Step 3: Calculate concentration of the solution
Concentration solution = moles NaOH / volume water
Concentration solution = 0.20 moles / 0.400 L
Concentration solution = 0.50 M
The concentration of this sodiumhydroxide solutions is 0.50 M
The ML of 0.85 m NaOH required to titrate 25 ml of 0.72m hbr to the equivalence point is calculated as follows
calculate the moles of HBr used
moles = molarity x volume
25 x0.072/1000= 0.0018 moles
write the equation for reaction
NaOH + HBr = NaBr + H2O
from reacting equation the mole ratio between NaOH to HBr is 1:1 therefore the moles of NaOH = 0.0018 moles
volume = moles/molarity
0.0018/0.085 = 0.021 L in Ml = 0.021 x1000=21.18 Ml ofNaOH
