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.
There are six atoms of oxygen in that compound.
It is a period.....
this ok??
Answer:
5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol.
Explanation:
The balanced reaction:
CO (g) + 2 H₂ (g) -> CH₃OH (l)
By stoichiometry of the reaction, the following amounts of moles of each compound participate in the reaction:
- CO: 1 mole
- H₂: 2 moles
- CH₃OH: 1 mole
Being the molar mass of each compound:
- CO: 28 g/mole
- H₂: 1 g/mole
- CH₃OH: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CO: 1 mole* 28 g/mole= 28 grams
- H₂: 2 moles* 1 g/mole= 2 grams
- CH₃OH: 1 mole* 32 g/mole= 32 grams
Being 6 kg equivalent to 6000 grams (1 kg= 1000 grams), you can apply the following rules of three:
- If by stoichiometry 32 grams of methanol are formed from 28 grams of carbon monoxide, 6000 grams of methanol are formed from how much mass of carbon monoxide?

mass of carbon monoxide= 5250 grams= 5.25 kg
If by stoichiometry 32 grams of methanol are formed from 2 grams of hydrogen, 6000 grams of methanol are formed from how much mass of hydrogen?

mass of hydrogen= 375 grams
<u><em>5250 grams or 5.25 kg of carbon monoxide and 375 grams of hydrogen are required to form 6 kg of methanol. </em></u>
Explanation:
S= 2
P=6
D=10
F= 14
F orbital holds a maximum of 14 electrons