This equation C5H + O2 ---> CO2 + H2O has a mistake.
C5H is wrong. You missed the subscript of H.
I will do it for you assuming some subscript to show you the procedure, but you have to use the right equation to get the right balanced equation.
Assuming the tha combustion equation is C5H12 + O2 ---> CO2 + H2O
First you need to balance C, so you put a 5 before CO2 and get
C5H12 + O2 ---> 5CO2 + H2O
Now you count the hydrogens: 12 on the left and 2 on the right. So put a 6 before H2O and get:
C5H12 + O2 ---> 5CO2 + 6H2O
Now count the oxygens: 2 on the left and 16 on the right, so put an 8 on before O2:
=> C5H12 + 8O2 ---> 5CO2 + 6H2O.
You can verify that the equation is balanced
Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
Hello.
If you wanted to harvest cranberries you would need to use flotation.
Have a nice day.
Answer:
I HOPE THE ABOVE INFORMATION WILL HELP YOU A LOT.
HAVE A NICE DAY.
Answer:
The coefficients are 6, 1, 3
Explanation:
HNCO →C3N3(NH2)3 + CO2
From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:
6HNCO → C3N3(NH2)3 + CO2
Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:
6HNCO → C3N3(NH2)3 + 3CO2
Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.
The coefficients are 6, 1, 3
