Answer:
5.83 mol.
Explanation:
- From the balanced reaction:
<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>
<em />
- The limiting reactant is Ag₂S.
- The excess "left over" reactant is Al.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.
Acacia ant and acacias animals are an example of coevolution
Coevolution is the reciprocal evolutionary change in a set of interacting population over time resulting from the interaction between those population and an example of coevolution that is not characteristics of an arm race but one which provides a mutual benefit to both a plant species and insect is that of the acacia ant and acacia plant and many cases of coevolution can be found between plants and insects
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Answer:
I think its B im not sure
but i hope this helps
Answer:
Here's what I get
Explanation:
1. Nickel sulfate
base + acid ⟶ salt + water
NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.
Hydroxides of transition metals are insoluble; most sulfates are soluble.
2. Carbonate + acid
Most carbonates are insoluble.
They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.
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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