help on chem.. ASAP?

Question

seawater is approximately 4.0 % by mass dissolved ions. about 85% of the mass of the dissolved ions is from NaCl.
a) calculate the mass percent of NaCl in seawter.
b) calculate the mass percent of Na+ ions and of Cl- ions in seawater.
c) calculate the molarity of NaCl in seawater at 15 Celcius degree ( d of seawater at 15 *C = 1.035 g/ml)

Answer 1

amu - Atomic mass units.

(a)
4% = 0∙04
85% = 0∙85
Mass % of NaCl = (0∙04)(0∙85)(100%) = 3∙4%

(b)
Na = 23 amu.
Cl = 35∙5 amu.
Total mass of mole of NaCl = 23 + 35∙5 = 58∙5 amu.
For Na: 23/58∙5 x 3∙4% = 1∙3367...%
For Cl: 35∙5/58∙5 x 3∙4% = 2∙0632...%

(c)
1∙035 g/ml
= 1035 g/L
3∙4% of 1035 g/L = 35∙19 g/L
1 Mole of NaCl = 58∙5g
Molarity of NaCl = 35∙19 / 58∙5 = 0∙6015...Moles.

Answer 2

You do do not forget that once a number of those chemical substances have expenses signifies that they are disassociated authentic? in hardship-free words sodium hydroxide NaOH, Na+ an OH- sodium bromide NaBr, Na+, Br- calcium oxide CaO, Ca 2+ , O 2- Carbon Monoxide CO, covalent bonds sulfur dioxide SO2, covalent bonds silver chloride AgCl, Ag+ Cl- copper hydroxide, Cu(OH)2, Cu2+, OH- zinc hydroxide Zn(OH)2 aluminum phosphate AlPO3 copper nitrate Cu(NO3)2 - Cu2+ NO3 - potassium carbonate K2CO3, ok+ CO3 2- lead chloride PbCl2, Pb2+, Cl- carbon dioxide CO2 - covalent those are those i will undergo in concepts off hand

Answer 3

a) 3.4% NaCl
Solution: 4.0% * 0.85 = 3.4%

b) 1.34% Na+, 2.06% Cl+
Solution: get the mass ratio of each atom from its molecular mass, and then multiply this to mass percent of NaCl.
For Na+, (23/58.5)*3.4% = 1.34%
For Cl-, (35.5/58.5)*3.4% = 2.06%

c) 0.601M
Solution: density times percent NaCl divided by molar mass of NaCl, and then convert mL to L. So,

(1.035 g/mL)* 0.034 * (1000mL/L) / (58.5 g/mol) = 0.601M