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ΔH = MCΔT
ΔH = to the amount of energy or change in energy (J)
mass of water
C = waters specific heat capacity
ΔT = change in temperature
and so ΔH = 25 × 4.18 × ( 112-67 ) J = 4702.5 J
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Answer: The kilograms of water must evaporate from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
Explanation:
According to the ratio and proportion:

where,
= concentration of ist solution = 25%
= mass of ist solution = 8 kg
= concentration of second solution = 40%
= mass of second solution = ? kg


Thus the final solution must have a mass of 5 kg , i.e (8-5)= 3 kg of mass must be evaporated.
Therefore, the mass that must be evaporated from 8kg of a 25% salt solution to produce 40% salt solution is 3 kg.
<span>The reason it will be 7 for some titrations is that when you titrates a strong acid with a strong base for example HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7
second reason is
When you titrates a weak acid with strong base at equivalence point
only a water solution of the conjugate base exists
CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
The complete balanced chemical reaction is written as:
AgNO3 + KCl ---> AgCl
+ KNO3
where AgCl is our
precipitate
So calculating for moles
of AgCl produced: MM AgCl = 143.5 g/mol
moles AgCl = 0.326 g /
(143.5 g/mol) = 2.27 x 10^-3 mol
we see that there is 1
mole of Ag per 1 mole of AgCl so:
moles Ag = 2.27 x 10^-3
mol
The molarity is simply
the ratio of number of moles over volume in Liters, therefore:
Molarity = 2.27 x 10^-3
mol / 0.0977 L
<span>Molarity = 0.0233 M</span>
Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
molarity of HCl=0.02117*10/1000
=2.117M