12 moles of water H₂O are produced from the combustion of pentane.
Explanation:
We have the following combustion of pentane (C₅H₁₂):
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
Knowing the chemical reaction we devise the following reasoning:
if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O
then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O
X = (2 × 6) / 1 = 12 moles of water H₂O
Learn more about:
combustion of organic compounds
brainly.com/question/7295137
brainly.com/question/884053
#learnwithBrainly
Answer:
You take the atomic, or proton number of the element, and you subtract it from the element's mass number.
D = m / V
13.6 = 8.3 / V
V = 8.3 / 13.6
V = 0.610 mL
hope this helps!
Answer:
0.106 mol (3s.f.)
Explanation:
To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.
Ar of C= 12, Ar of H= 1, Mr of O= 16
These Ar values can be found on the periodic table.
Mr of glucose= 6(12)+ 12(1) + 6(16)= 180
Moles of glucose
= mass ÷ mr
= 19.1 ÷ 180
= 0.106 mol (3 s.f.)
the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
Learn more about Arrhenius equation here-
brainly.com/question/12907018
#SPJ4
