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
.
Meiosis 1 is the reductional division ! so after telophase 1 , cytokinesis takes place and two haploid cells are formed !
Answer:
(C) through the atmosphere
Explanation:
1 gallon of antifreeze = 60% of mixture. Total mixture:
Vm = 1/0.6
= 1.67 gallons
Volume of water = total vol - antifreeze vol
= 1.67 - 1
= 0.67 gallon of water