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san4es73 [151]
2 years ago
8

Help me┐(´ー`)┌thank you in advance ​

Chemistry
2 answers:
LenKa [72]2 years ago
8 0

<u>Answer</u>:

the law of conservation of mass is valid for all the following, except nuclear reactions.

<u>Explanation</u>:

the law of conservation of mass is valid for all the following, except nuclear reactions because some mass gets converted to energy.

liubo4ka [24]2 years ago
3 0

Answer:

Nuclear reactions

Explanation:

What's this law?

  • This law states that in any chemical reaction the mass is neither created nor destroyed ,it remains conserved
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Vinil7 [7]

Answer:

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6 0
3 years ago
A West Virginia coal is burned at a rate of 8.02 kg/s. The coal has a sulfur content of 4.40 % and the bottom ash contains 2.80
aleksley [76]

Answer: The annual emission rate of SO2 is 1.08 × 10^{7} 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.

5 0
3 years ago
Typical frequencies for several types of electromagnetic radiation are given below. Calculate the energy of each type of radiati
marusya05 [52]

Explanation:

The energy of electromagnetic radiation is given by :

E=h\nu

Where, h is the Planck's constant

(a) A gamma ray, \nu = 1.64\times 10^{20}\ Hz

Energy,

E=6.63\times 10^{-34}\times 1.64\times 10^{20}\\E=1.08\times 10^{-13}\ J

(b) A microwave, \nu = 4.95\times 10^{10}\ Hz

Energy,

E=6.63\times 10^{-34}\times 4.95\times 10^{10}\\E=3.28\times 10^{-23}\ J

(c) A AM radio wave, \nu = 2\times 10^{6}\ Hz

Energy,

E=6.63\times 10^{-34}\times 2\times 10^{6}\\E=1.32\times 10^{-27}\ J

Therefore, this is the required solution.

4 0
3 years ago
During and exothermic reaction, energy is _______ the surroundings.
Kitty [74]
Your answer is A mixed with.

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