1. What volume in mL of 28.0M H2SO4 is needed to contain 53.0g of H2SO4?

2. How many grams of calcium hydroxide Ca(OH)2 are needed to make 532.0 mL of a 1.90M solution?

3. If 12 mL of water are added to 124mL of a 0.505M K2SO4 solution, what will the molarity of the diluted solution be?

1. V=19.3 ml

2. mass=74.8g

3. molarity of dilted solution=0.463M

Explanation:

1.

First calculate the mole ,

given mass=53g

molecular weight=98g/mole

mole=0.54

molarity=28M

V=19.285ml

V=19.3 ml

2.

First calculate the mole of calcium hydroxide,

molariy=1.9M

volume=532 ml=0.532L

mole=1.01 mol;

molecular weight of calcium hydroxide= 74g/mol

mass=74.8g

3.

First calculate the mole before adding water

molariy=0.505M

initial volume=124 ml=0.124L

mole=0.063 mol;

after dilution volume=136 ml =0.136L

quantity of solute remains unchanged during dilution of solution.

molarity of dilted solution=0.463M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

2b2t

explanation:

2b2t