Mass of solute = 25.8 g
mass solution = 212 g
% =( mass of solute / mass solution ) x 100
% = ( 25.8 / 212 ) x 100
% = 0.122 x 100
= 12.2 %
Answer:
146.3g NaCl (mol NaCl/58.44g NaCl) = 2.50 mol NaCl
1.5M NaCl = 1.5 mol NaCl / 1 L = 2.5 mol NaCl / x L, solve for x
x L = 2.5 mol NaCl / 1.5 mol NaCl = 1.66 L
It gives the answer and all the working.
To put it another way:
Dividing the amount required by the molar mass
we quickly see that 2.5 moles are required.
One litre of 1.5 molar solution gives 1.5 moles
we need a further mole, which is 2/3 of 1.5 so 2/3 of a litre.
Answer:
q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.
Explanation:
The Specific Heat capacity of Lead is 0.128 
This means, increase in temperature of 1 gm of lead by
will require 0.128 J of heat.
Formula Used :

q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
= Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -
put value in formula
q = - 
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process