Answer:
Molar percent of sodium in original mixture is 88,50%
Explanation:
The last reaction is:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
The moles of BaCl₂ are:
0,132L × 3,80M = 0,502 moles of BaCl₂
As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.
0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄
These moles of Na₂SO₄ comes from:
2 Na + H₂SO₄ → Na₂SO₄ + H₂
As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄
0,502 moles of Na₂SO₄ ×
× 22,99 g/mole = 23,08 g of Na
Molar percent of sodium in original mixture is:
= <em>88,50% </em>
I hope it helps
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Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react
