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need help with chem AS

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ksitna:
17.1 g of aluminium sulfate, Al2(SO4)3, was dissolved in water.
Calculate the number of sulfate ions, SO4
2–, present in the solution formed.
[Assume the molar mass of Al2(SO4)3 is 342 g mol–1 and the Avogadro Constant is
6 × 1023 mol–1.]

how do u do this
can u help me pleassee

01chris:
The formula of the compound Al2(SO4)3 tells you that you have 3 SO4 ions for every molecule of the compound.

You need to work out how many moles of the compound you have: n = m/M  -or as I like to remember it- no. moles = grams/RAM.

17.1g/342g mol-1 = 0.05mol.

Avagadro's number says that a mole of a substance = 6x1023 particles. You have 0.05 mol therefore 0.05 x 6x1023 particles (which equals 3x1022 ).

Going back to what we said at the start: for every Al2(SO4)3 molecule you have 3 SO4 ions. You have 3x1022 Al2(SO4)3 therefore you have 3 x 3x1022 SO4 ions (= 9x1022 ).

Does that help?

Dento:
hey ksitna,
here's how to deal with these questions,

first, find the numer of moles of the sulphate ions which is >>>> 17.1/342 = 0.05 moles

then multply this mole number with the avogadro's constant and 3.    ( 3 is the number of sulphates given in the question >> Al2(SO4)"3" )

so, it goes this way,



moles of sulphate = 0.05 moles
number of sulphate ions present = 0.05 x (6x10 raised 23) x 3
= 9x10 raised 22
Answer is "D"

this is the final answer, hope i helped u, if there is anything else, feel free to ask
good luck this thursday !!

ksitna:
THANK UUU!!!!!!!
I OWE U GUYS ONE
THANKS A LOOTTTT ;D

ksitna:
sooo sorry to be a bother
AGAIN
but um are the oxidation stuff and flame tests required for this module?

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