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.
Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
The number of grams of aluminum required to react with HCl is 0.6258 g.
Hence, concentration of base is 1.17 M
Answer:
Density = 11.4 g/cm³
Explanation:
Given data:
Density of lead = ?
Height of lead bar = 0.500 cm
Width of lead bar = 1.55 cm
Length of lead bar = 25.00 cm
Mass of lead bar = 220.9 g
Solution:
Density = mass/ volume
Volume of bar = length × width × height
Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm
Volume of bar = 19.4 cm³
Density of bar:
Density = 220.9 g/ 19.4 cm³
Density = 11.4 g/cm³
