# chemistry

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A laboratory procedure calls for making 400.0 mL of a 1.1 M NnO3 solution. What mass of NaNO3 (in g) is needed?

• chemistry -

M = moles/L
Solve for moles.

moles = grams/molar mass
solve for grams.

• chemistry -

I dunno......I hate chemistry!

• chemistry -

• chemistry -

All you need to do is to substitute your problem numbers into the equations I've given you.
M = moles/L
1.1 = moles/0.400 L
Solve for moles, the only unknown.
moles = 1.1 x 0.400 = 0.440 mols.

Second equation:
moles = grams/molar mass
moles from above = 0.440
molar mass NaNO3 is about 85 (23+14+3*16) = about 85 but you need to confirm that.
So 0.440 = grams/85
Solve for grams, the only unknown.
g = 0.440 x 85 =37.4 grams NaNO3 in 400 mL (0.400 L).

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