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Nadya [2.5K]
4 years ago
12

Many organisms in Kingdom Fungi grow best in warm, moist conditions.

Chemistry
1 answer:
almond37 [142]4 years ago
3 0

Answer:

Kingdom Fungi refers to the group of eukaryotic organisms like fungi, mushrooms, molds and yeasts.

Explanation:

The fungi kingdom is made up of eukaryotic organisms whose cell walls are made of chitin.

Next I will name <u>some characteristics</u> of these organisms:

  • They are not plants or animals .
  • They are responsible for absorbing nutrients from other organisms, which they use as food .
  • Some, like mushrooms and yeast, are edible.
  • They present cellular orguils and vacuoles .
  • They lack chloroplasts .
  • They produce spores .
  • They are able to break down dead matter .
  • They develop in humid and dark areas .
  • Penicillin is obtained from these organisms.
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A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
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Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

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