Answer:
Al2(SO4)3 and Mg(OH)2
Explanation:
1. Al has a charge of 3-, and SO4 of 2-
when you cross multiply the charges you get
Al2 and (SO4)3
*the reason theres a bracket around the sulfate ion is that the charge 3 is not for oxygen only, but the entire sulphate ion*
Hence, Al2(SO4)3
2. Mg has a charge of 2- and OH of 1-
again cross multiply
Mg (you dont need to add the 1) and (OH)2
again, the bracket around OH means the charge appiles to Oxygen AND hydrogen
hence, Mg(OH)2
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.
pls mark as brainliest ans
1) Formulas:
a) mole fraction of component 1, X1
X1 = number of moles of compoent 1 / total number of moles
b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass
2) Application
Number of moles of CaI2 = 0.400
Molar mass of water = 18.0 g/mol
Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol
Total number of moles = 0.400 + 47.22 =47.62
Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
Yes, Benzylamine is miscible, meaning it is soluble at all amounts.
Answer:
E
Explanation:
A catalyst is a substance which alters the rate of a chemical reaction. We check for the correctness of each of the options as follows:
A. Is wrong.
A catalyst can increase the rate of the forward and backward reaction
B. Is wrong
A catalyst does not slow the reverse reaction only. This particular case is the case of a negative catalyst
C is wrong
A catalyst has no effect on the equilibrium nor the equilibrium constant
D is wrong
Catalyst has no effect on equilibrium value
E is correct
Although Catalysts has no effect on equilibrium or its constant value, it can increase the the rate at which equilibrium is achieved by speeding up the reaction through bringing down the activation energy
