Answer:
hello your question is incomplete below is the missing part of the question
answer : 104°c
Explanation:
The Eutectic temperature for the mixture is 104°c
From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c
The solution to the problem is as follows:
<span>a)
C + O2 = CO2
Molar mass CO2 = 44 g/mol
3.67 g CO2 * 1 mol / 44 g =
=0.0834 mol CO2 = 0.0834 mol C
I hope my answer has come to your help. God bless and have a nice day ahead!
</span>
Answer:
D
Explanation:
To answer this question, we will need to write the dissociation equation of aluminum trichloride.
AlCl3 ——-> Al3+ + 3Cl-
It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion.
From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J
(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal
(rounded to three significant figures)
95.6 cal
are needed.
