Answer:
a. 50ml b.10ml c. 6.097ml d. 190.1 ml
Explanation:
According to Boyle's law
Volume is inversely proportional to pressure at constant temerature
Mathematically
P1V1=P2V2
P1=Initial pressure=0.8atm
V1=Initial volume=25ml
making V2 the subject
at 0.4atm P2=0.4 atm,
V2=25×0.8/0.4
=50ml
at 2 atm V2=25×0.8/2
=10 ml
1mmHg=0.00131579
2500mmHg=3.28 atm
At 3.28 atm,V2=25×0.8/3.28
=6.097 ml
at 80.0 torr
1 torr=0.00131579
80 torr=0.1052 atm
at 0.1048 atm V2=25×0.8/0.1048
=190.1 ml
Sunlight i think thats the answer
Answer:
The molarity of the dissolved NaCl is 6.93 M
Explanation:
Step 1: Data given
Mass of NaCl = 100.0 grams
Volume of water = 100.0 mL = 0.1 L
Remaining mass NaCl = 59.5 grams
Molar mass NaCl= 58.44 g/mol
Step 2: Calculate the dissolved mass of NaCl
100 - 59. 5 = 40.5 grams
Step 3: Calculate moles
Moles NaCl = 40.5 grams / 58.44 g/mol
Moles NaCl = 0.693 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity dissolved NaCl = 0.693 moles / 0.1 L
Molarity dissolved NaCl = 6.93 M
The molarity of the dissolved NaCl is 6.93 M
Answer: 1. 
moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 
moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove 
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide= 
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .
