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.
Answer:
-3.28 × 10⁴ J
Explanation:
Step 1: Given data
- Pressure exerted (P): 27.0 atm
- Initial volume (Vi): 88.0 L
- Final volume (Vf): 100.0 L
Step 2: Calculate the work (w) done by the gaseous mixture
We will use the following expression.
w = -P × ΔV = -P × (Vf - Vi)
w = -27.0 atm × (100.0 L - 88.0 L)
w = -324 atm.L
Step 3: Convert w to Joule (SI unit)
We will use the conversion factor 1 atm.L = 101.325 J.
-324 atm.L × 101.325 J/1 atm.L = -3.28 × 10⁴ J
Total number of atoms = 7
Total number of H atom = 5
% of H in ammonium hydroxide = 5/7 ×100 = 71.4 %
Answer:
A. Solution, Colloid, Suspension
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