Answer:
The activation energy is 7.11 × 10⁴ J/mol.
Explanation:
Let's consider the Arrhenius equation.
where,
k is the rate constant
A is a collision factor
Ea is the activation energy
R is the ideal gas constant
T is the absolute temperature
The plot of ln k vs 1/T is a straight line with lnA as intercept and -Ea/R as slope. Then,
Answer:
ammonia
Explanation:
Nitrogen fertilizers contain N in the forms of ammonium, nitrate and urea. Upon application to the soil, urea-N rapidly hydrolyzes to ammonia, thus it shares similar characteristics as ammonia-based N fertilizers.
Metals react with ______ to form compounds that are alkaline.
A. metalloids
B. oxygen (O)
C. non-metals
D. hydrogen (H)
The answer is D, Hydrogen (H).
The mixture flow rate in lbm/h = 117.65 lbm/h
<h3>Further explanation</h3>
Given
15.0 wt% methanol
The flow rate of the methyl acetate :100 lbm/h
Required
the mixture flow rate in lbm/h
Solution
mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :
mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :
Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.
1 kg mixture = 0.85 .methyl acetate
So flow rate for mixture :
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
