Explanation:
Moles of metal,
=
4.86
⋅
g
24.305
⋅
g
⋅
m
o
l
−
1
=
0.200
m
o
l
.
Moles of
H
C
l
=
100
⋅
c
m
−
3
×
2.00
⋅
m
o
l
⋅
d
m
−
3
=
0.200
m
o
l
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if
0.200
m
o
l
acid react, then (by the stoichiometry), 1/2 this quantity, i.e.
0.100
m
o
l
of dihydrogen will evolve.
So,
0.100
m
o
l
dihydrogen are evolved; this has a mass of
0.100
⋅
m
o
l
×
2.00
⋅
g
⋅
m
o
l
−
1
=
?
?
g
.
If 1 mol dihydrogen gas occupies
24.5
d
m
3
at room temperature and pressure, what will be the VOLUME of gas evolved?
Glucose is the starting molecule for glycolysis.
Answer:
T2 = 51.6°C
Explanation:
Given:
P1 = 1.01 atm
T1 = 25°C + 273 = 298K
P2 = 1.10 atm
T2 = ?
P1/T1 = P2/T2
Solving for T2,
T2 = (P2/P1)T1
= (1.10 atm/1.01 atm)(298K)
= 324.6 K
= 51.6°C
where Tc = Tk - 273
Answer:
A
Explanation:
What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3
3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply
3x = 2 * 300
3x = 600 Divide by 3
3x/3 = 600/3 Do the division
x = 200.00
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
