12 % salt is present in 125 g mixture of salt and sand.
Keep in mind that the total percentage is always 100 %
Therefore, if 12 % is the salt, remaining 88 % must be
sand.
a. The amount of mixture is 125 q. Here, 12 % of 125 is
12 * 125 / 100 = 15 g of salt is present in 125 g mixture.
b. The amount of sand can be calculated similarly, 88 %
of 125 g is 88 * 125 / 100 = 110 g of sand is present in
125 g mixture.
Answer:
amount of silver chloride required is 0.015 moles or 2.1504 g
Explanation:
0.1M AgCL means 0.1mol/dm³ or 0.1mol/L
1L = 1000mL
if 0.1mol of AgCl is contained in 1000mL of solution
then x will be contained in 150mL of solution
cross multiply to find x
x = (0.1*150)/1000
x= 0.015 moles
moles of silver chloride present in 150 mL of solution is 0.15 moles
To convert this to grams, simply multiply this value by the molar mass of silver chloride
molar mass of silver chloride AgCl =107.86 + 35.5
=143.36 g/mol
mass of AgCl = moles *molar mass
=0.015*143.36
=2.1504g
=
So you need to put numbers before each compound to make sure there are the exact same number of elements on each side. If you put a 4 before NH4 there are 4 Nitrogen and now 16 hydrogen. I just played around with numbers and guessed until I got them even.
energies of electrons can only have certain specific values.