This process is called aerobic respiration.
The question is incomplete, the complete question is;
Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.
Answer:
AlCl3 0.100 m
Explanation:
Let us remember that the boiling point elevation is given by;
ΔTb = Kb m i
Where;
ΔTb = boiling point elevation
Kb = boiling point constant
m = molality of the solution
i = Van't Hoff factor
We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.
AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.
A. Petrified fossil should be the answer
Answer:
The volume of 5.0 g CO 2 is 2.6 L CO 2 at STP
Explanation:
STP
STP is currently
0
∘
C
or
273.15 K
, which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is
10
5
.
Pascals (Pa)
, but most people use
100 kPa
, which is equal to
10
5
.
Pa
.
You will use the ideal gas law to answer this question. Its formula is:
P
V
=
n
R
T
,
where
P
is pressure,
V
is volume,
n
is moles,
R
is a gas constant, and
T
is temperature in Kelvins.
Determine moles
You may have noticed that the equation requires moles
(
n
)
, but you have been given the mass of
CO
2
. To determine moles, you multiply the given mass by the inverse of the molar mass of
CO
2
, which is
44.009 g/mol
.
5.0
g CO
2
×
1
mol CO
2
44.009
g CO
2
=
0.1136 mol CO
2
Organize your data
.
Given/Known
P
=
100 kPa
n
=
0.1136 mol
R
=
8.3145 L kPa K
−
1
mol
−
1
https://en.wikipedia.org/wiki/Gas_constant
T
=
273.15 K
Unknown:
V
Solve for volume using the ideal gas law.
Rearrange the formula to isolate
V
. Insert your data into the equation and solve.
V
=
n
R
T
P
V
=
0.1136
mol
×
8.3145
.
L
kPa
K
−
1
mol
−
1
×
273.15
K
100
kPa
=
2.6 L CO
2
rounded to two significant figures due to
5.0 g
Answer link
Doc048
May 18, 2017
I got 2.55 Liters
Explanation:
1 mole of any gas at STP = 22.4 Liters
5
g
C
O
2
(
g
)
=
5
g
44
(
g
mole
)
=
0.114
mole
C
O
2
(
g
)
Volume of 0.114 mole
C
O
2
(
g
)
= (0.114 mole)(22.4 L/mole) = 2.55 Liters
C
O
2
(g) at STP
