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daser333 [38]
3 years ago
14

The label on a tub of margarine states that 100 g of the margarine contains 0·70 g of sodium. The sodium is present as sodium ch

loride. Calculate the mass of sodium chloride, in grams, present in a 10 g portion of the margarine. The mass of one mole of sodium chloride, NaCl, is 58·5 g.
Chemistry
1 answer:
Naddika [18.5K]3 years ago
7 0
Here you have one tenth of the original quantity of margarine.
if 100 grams of margarine has 0.70 grams of sodium, then one tenth of the margarine has one tenth of the sodium. So 10 grams of margarine has
0.070 grams of sodium chloride, keeping track of the significant figures.
If, for some reason, you need the number of moles present in 10 grams, it would be:
(#grams sodium chloride)/(#grams/mole of sodium chloride)
(0.070 grams)/(58.5 grams/mole) =  0.0012 moles of Sodium Chloride
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What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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