Joshua dissolved sugar in water to create a sugar solution. His friend Ray tells him it’s a compound, but Joshua corrects him. W
2 answers:
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Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M 30 / 1000 L = 0.140 M
0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol ( 1 L / 0.570 mol )
( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer : The enthalpy of reaction is, 67.716 KJ/mole
Explanation :
First we have to calculate the moles of and
.
Now we have to calculate the moles of AgCl formed.
The balanced chemical reaction will be,
As, 1 mole of react with 1 mole of
to give 1 mole of
So, 0.005 mole of react with 0.005 mole of
to give 1 mole of
The moles of AgCl formed = 0.005 mole
Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml
Now we have to calculate the mass of solution.
Mass of the solution = Density of the solution × Volume of the solution
Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g
Now we have to calculate the heat.
where,
q = heat
C = specific heat capacity =
m = mass = 100 g
= final temperature =
= initial temperature =
Now put all the given values in the above expression, we get:
Now we have to calculate the enthalpy of the reaction.
where,
= enthalpy of reaction = ?
q = heat of reaction = 338.58 J
n = moles of reaction = 0.005 mole
Now put all the given values in above expression, we get:
Conversion used : (1 KJ = 1000 J)
Therefore, the enthalpy of reaction is, 67.716 KJ/mole