Oxidation of D-glucose are usually found on the photosynthesis which has a by product of carbon dioxide (CO2) and water (H2O) chemically written as,
C6H12O6 +6O2----> 6CO2 +6H2O
Moreover, D-glucose can also be chemically react with oxygen to form D-gluconic acid and D-glucoronic acid.
Answer:
This question is incomplete
Explanation:
This question is incomplete because of the absence of options. However, the compound C₆H₁₄ is hexane. Hexane is a member of saturated hydrocarbons (homologous series) called alkanes (with the general formula CₙH₂ₙ₊₂). The structure for an hexane is shown below
H H H H H H
I I I I I I
H - C - C - C - C - C - C - H
I I I I I I
H H H H H H
which can also be written as
CH₃CH₂CH₂CH₂CH₂CH₃
Using ideal gas equation,
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=1 atm
T=25 C+273 K =298.15K
V=663 ml=0.663L
R=0.0821 atm L mol ⁻¹
Mass of gas given=1.25 g g
Molar mass of gas given=?
Putting all the values in the above equation,
Molar mass of the gas=46.15
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Explanation:
Equation for the given reaction is as follows.
Therefore, moles of NaOH and HA are calculated as follows.
Moles of NaOH = 
= 
= 0.00864 mol
Moles of HA = 0.00864
Also, moles = 
Molecular weight = 
= 122.22 g/mol
Thus, we can conclude that molar mass of given unknown organic acid is 122.22 g/mol.
