Answer:
The amount of CaCl2 produced depends on the amount of HCl in the reaction.
Explanation:
The amount of HCl is used completelyin the reaction unlike CaCO3 which remains after reaction.
V ( HCl ) = 16.4 mL / 1000 => 0.0164 L
M( HCl) = ?
V( KOH) = 12.7 mL / 1000 => 0.0127 L
M(KOH) = 0.620 M
Number of moles KOH:
n = M x V
n = 0.620 x 0.0127
n = 0.007874 moles of KOH
number of moles HCl :
<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
</span>
moles HCl = 0.007874 * 1 / 1
= 0.007874 moles of HCl
M = n / V
M = 0.007874 / <span>0.0164
</span>= 0.480 M
Answer (2)
hope this helps!
The term formula units means molecules.
Then, what you are looking for is the mass in 4.59*10^24 molecules.
The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.
1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol
2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
3) mass of NaCl = molar mass * number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl
Answer: 445.31 g of NaCl.
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
