Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y
3y = 1.2
y = 0,4M
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4
C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄
mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol
164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
Answer:
A hypothesis or a model is called falsifiable if it is possible to conceive of an experimental observation that disproves the idea in question.
<span>Data:
1) molar
mass: 58.0714 g.
2) mass analysis
C: 18 g
H: 3 g
O: 8 g.
Solution
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<span>What is the percent composition of all the components in the liquid?
mass of C + mass of H + mass of O = 18g + 3g + 8g = 29 g
Percent composition:
C: (18g / 29g) * 100 = 62%
H: (3g / 29g) * 100 = 10%
O: (8g / 29g) * 100 = 28%
--------------
Total 100%
What is the empirical formula of the compound?
Convert masses to number of moles by dividing each mass by the atomic mass of the correspondant element:
C: 18 g/ 12.0 g/mol = 1.5 mol
H: 3 g / 1.0 g/mol = 3 mol
O: 8 g / 16 g/mol = 0.5 mol
Divide each number of moles by the smaller, i.e. 0.5 mol
=> C: 1.5 mol / 0.5 mol = 3 mol
=> H: 3 mol / 0.5 mol = 6 mol
=> O: 0.5 mol / o.5 mol = 1 mol
=> Empirical formula: C3 H6 O
What is the molecular formula of the unknown liquid?
First,calculate the molar mass of the empirical formula:
C3 H6 O => 3 * 12 g/mol + 6*1 g/mol + 16 g/mol = 58 g/ mol
Secont, given that the molar mass is the same that the empirical mass, the molecular formula is the same empirical formula.
=> Molecular formula = C3 H6 O.
This compound has been
identified as a ketone.
Go online and figure out the name of this
compound.
It is easy to find that this compound is propanone or acetone (both names are valid).
Now you can go online and find its characteristics and common uses.
Can you identify
the killer from this information? How?
Likely they think you can presume that the killer is the actress, because women use acetone regularlty to remove the nail polish.
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