The moles of butane gas and oxygen gas reacted if 2.50 moles of H2O is produced is calculated as below
the equation for reaction
2C4H10 +13 O2 = 8CO2 +10 H2O
the moles of butane (C4H10) reacted calculation
by use of mole ratio between C4H10: H2O which is 2 : 10 the moles of C4H10= 2.50 x2/10 = 0.5 moles of C4H10 reacted
The moles of O2 reacted calculation
by use of mole ratio between O2 : H2O which is 13:10 the moles of O2
= 2.50 x 13/10= 3.25 moles of O2 reacted
Answer:
We will derive the combined gas equation from the law of Gases
Boyle's law
P∝1/V
PV=Constant
Charles law
V∝T
Avogadro's Law
V∝n
By combining these three equations we get the combined gas equation
PV=nRT
V=nRT/P
If n=1 mole
V=RT/P
By putting the value of R, T, and P in the above equation we can calculate the volume of the gas at STP.
Answer:
Non competitive inhibition
Explanation:
Hello,
During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.
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I think it's less, more, more. correct me if i'm wrong
