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.
C- Sugar is very much soluble (example- putting sugar in tea), sand is definately not soluble (example- a sandy beach) and cornstarch is the last one, so logically it should be somewhat soluble.
Answer:
B. choosing a material that will show warning before it fails.
Explanation:
Answer:
The answer to your question is 122.4 g of O₂
Explanation:
Data
mass of O₂ = ?
moles of H₂O = 7.65
Process
1.- Write the balanced chemical reaction
2H₂O ⇒ 2H₂ + O₂
2.- Convert the moles of H₂O to grams
molar mass of H₂O = 2 + 16 = 18 g
18 g of H₂O ---------------- 1 mol
x ----------------- 7.65 moles
x = (7.65 x 18) / 1
x = 137.7 g H₂O
3.- Calculate the grams of O₂
36 g of H₂O -------------------- 32 g of O₂
137.7 g of H₂O ------------------- x
x = (32 x 137.7) / 36
x = 122.4 g of O₂
Answer :
Example of polar covalent molecules H-O-H(water), ammonia
Explanation:
The presence of intermolecular Hydrogen bonding makes the boiling point of water unexpectedly high, and the polar covalent nature makes it dissolve polar solute/compound
