This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
Answer:
<h2>Density = 0.00026 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass of air = 1.2 g
volume = 4,555 mL
Substitute the values into the above formula and solve for the density
That's
<h3>
</h3>
= 0.0002634
We have the final answer as
<h3>Density = 0.00026 g/mL</h3>
Hope this helps you
Answer:
0.64 L
Explanation:
Recall that
n= CV where n=m/M
Hence:
m/M= CV
m= given mass of solute =152g
M= molar mass of solute
C= concentration of solute in molL-1 = 1.5M
V= volume of solute =????
Molar mass of potassium permanganate= 158.034 g/mol
Thus;
152 g/158.034 gmol-1= 1.5M × V
V= 0.96/1.5
V= 0.64 L
The formula unit for Li and O is —————-> Li2O
