Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium concentration of
= 
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
For the given chemical reaction:
The expression for
is written as:
Thus the value of the equilibrium constant Kc for this reaction is 3.72
Answer:
60 grams
Explanation:
We have the balanced equation (without state symbols):
6
H
2
O
+
6
C
O
2
→
C
6
H
12
O
6
+
6
O
2
So, we would need six moles of carbon dioxide to fully produce one mole of glucose.
Here, we got
88
g
of carbon dioxide, and we need to convert it into moles.
Carbon dioxide has a molar mass of
44
g/mol
. So here, there exist
88
g
44
g
/mol
=
2
mol
Since there are two moles of
C
O
2
, we can produce
2
6
⋅
1
=
1
3
moles of glucose
(
C
6
H
12
O
6
)
.
We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.
Glucose has a molar mass of
180.156
g/mol
. So here, the mass of glucose produced is
1
3
mol
⋅
180.156
g
mol
≈
60
g
to the nearest whole number.
So, approximately
60
grams of glucose will be produced.
A.Weak
B.concentrated
C.dilute
D.strong
Answer: A - Weak
274ml's that is if you are converting from Liters