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:
A. their cells contain a nucleus
Explanation:
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Answer:
Protons and Neutrons
Explanation:
The sum of number of protons and the number of neutrons is equal to the mass of an element. Mathematically, it can be given as :
M = P + N
Also, the atomic mass can be calculated by the sum of the element's isotopes as isotopes have slightly different mass numbers.
Hence, protons and neutrons add together to determine the atomic mass.
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of fusion = 3.17 kJ/mol
As we know that:
= freezing point temperature = 
Now put all the given values in the above formula, we get:
Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
