Answer:
3Na2SO4
Explanation:
3Na2SO4
# of molecules: 3 moles of Na2SO4 or 3 × 6.22 × 10^23 molecules.
# of elements: 3 elements l
Name of element: = Sodium, S = Sulphur, O = Oxygen
# of atoms: Na = 6 atoms, S= 3, O= 12
Total # of atoms: 21
The #3 is a Coefficient.
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Hydrogen is composed of H atom and oxygen is composed of O atom. For water, it is composed by both H and O atom. If you burn hydrogen in oxygen, you can get water. And if you electrolysis water, you can get hydrogen and oxygen.
LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!
Answer:
Explanation:
This is a limiting reactant problem.
Mg(s)
+
2HCl(aq)
→
MgCl
2
(
aq
)
+ H
2
(
g
)
Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).
4.86
g Mg
×
1
mol Mg
24.3050
g Mg
=
0.200 mol Mg
Determine Moles of 2M Hydrochloric Acid
Convert
100 cm
3
to
100 mL
and then to
0.1 L
.
1 dm
3
=
1 L
Convert
2.00 mol/dm
3
to
2.00 mol/L
Multiply
0.1
L
times
2.00 mol/L
.
100
cm
3
×
1
mL
1
cm
3
×
1
L
1000
mL
=
0.1 L HCl
2.00 mol/dm
3
=
2.00 mol/L
0.1
L
×
2.00
mol
1
L
=
0.200 mol HCl
Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,
2.01588 g/mol
0.200
mol Mg
×
1
mol H
2
1
mol Mg
×
2.01588
g H
2
1
mol H
2
=
0.403 g H
2
0.200
mol HCl
×
1
mol H
2
2
mol HCl
×
2.01588
g H
2
1
mol H
2
=
0.202 g H
2
The limiting reactant is
HCl
, which will produce
0.202 g H
2
under the stated conditions.
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