Answer:
C
Explanation:
Because that's the spin in its orbitals.
Answer: The annual emission rate of SO2 is 1.08 ×
kg/yr
Explanation:
- The rate <em>r</em> at which the coal is been burnt is 8.02 kg/s.
- Amount of sulphur in the burning coal is given as 4.40 %
i.e., 4.4/100 × 8.02 = 0.353 kg/s. Which is equivalent to the rate at which the sulphur is been burnt.
- Since the burning of sulphur oxidizes it to produce SO2, it follows that the non-oxidized portion of the sulphur will go with the bottom ash.
- The bottom ash is said to contain 2.80 % of the input sulphur.
- Hence the portion of the SO2 produced is 100 — 2.80 = 97.20 %.
- The rate of the SO2 produced is percentage of SO2 × rate at high sulphur is been burnt.
= 97.20/100 × 0.353 kg/s.
= 0.343 kg/s.
- To get the annual emission rate of SO2, we convert the kg/s into kg/yr.
1 kg/s = 1 kg/s × (60 × 60 × 24 × 365) s/yr
1 kg/s = 31536000 kg/yr
- Therefore, 0.343 kg/s = 0.343 × 31536000 kg/yr
= 10816848 kg/yr
= 1.08 × 10^7 kg/yryr.
If we consider a combustion reaction of Methane:
The balanced equation is:
CH4 + 2O2 ---> 2H2O + CO2
The rate of appearance of H2O is rH2O, rate of disappearance of O2 is -rO2
(rH2O)^2 = (-rO2)^2
rH2O = -rO2
Answer:
Explanation:
19) it is 3d10 instead of 4d10
20) it is missing 3p6, and 4s2 before 3d5
21) Ra is not a noble gas
22) Cs is not a noble gas