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.
if i am correct it shall be 12. because i am thinking, 1 mole = 1 ar.
Answer:
What do you need help with?
Explanation:
It either forms a base or an acid...I think that's the answer you're looking for
Explanation:
A.
In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.
B.
Equation of the reaction
2NaOH + H2SO4 --> Na2SO4 + 2H2O
Number of moles of H2SO4 = molar concentration × volume
= 0.75 × 0.0105
= 0.007875 moles.
By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875
= 0.01575 moles.
Molar concentration of NaOH = number of moles ÷ volume
= 0.01575 ÷ 0.0175
= 0.9 M of NaOH.
