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:
5 × 10^-4 L
Explanation:
The equation of the reaction is;
2KClO3 = 2KCl + 3O2
Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles
From the stoichiometry of the reaction;
2 moles of KClO3 yields 3 moles of O2
0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas
From the ideal gas equation;
PV= nRT
P= 85.4 × 10^4 KPa
V=?
n= 0.165
R= 8.314 J K-1 mol-1
T= 40+273 = 313K
V= 0.165 ×8.134 × 313/85.4 × 10^4
V=429.4/85.4 × 10^4
V= 5 × 10^-4 L
Sorry bro but idk maybe its :1 L=<span>103</span><span>mL</span>
answer:
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explanation:
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