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Hunter-Best [27]
3 years ago
12

A scientist eliminated all living organisms in a jar of soup by heating it. He then left the soup jar uncovered for some days. A

fter a few days, he saw molds in the soup. What most likely formed the mold?
Chemistry
2 answers:
velikii [3]3 years ago
6 0
Spores in the air or contamination through some sort of insect interaction. In an open system, there are billions of microbial agents floating around. With an open container containing a food source and time, the likelihood of contamination is nearly %100.
tigry1 [53]3 years ago
6 0

Answer:

Spores transported by the air.

Explanation:

Hello,

In biology, molds account for masses of mycelium, that is vegetative filaments or hyphaes and fruiting structures, produced by various fungi like Aspergillus, Penicillium, and Rhizopus. Such fungi, are associated with food spoilage and plant diseases, therefore, when they reproduce, they release spores which are spread to the air and subsequently transported to suitable sources of substrate allowing them to grow again. In such a way, since the jar was uncovered, spores transported by the air were allowed to get into the soup and grow there.

Best regards.

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17.6

Explanation:

Answer is explained above

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Cobalt-62 is a radionuclide with a half life of 1.5 minutes. What fraction of
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Calculate the percent ionic, the percent covalent, and the bond length (in picometers) of a chemical bond between phosphorus and
kirza4 [7]

Answer:

The correct option is;

4 percent ionic, 96 percent covalent, 222 pm

Explanation:

The parameters given are;

Phosphorus:

Atomic radius = 109 pm

Covalent radius = 106 pm

Ionic radius = 212 pm

Electronegativity of phosphorus = 2.19  

Selenium:

Atomic radius = 122 pm

Covalent radius = 116 pm

Ionic radius = 198 pm

Electronegativity of selenium= 2.55  

The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;

Using Pauling's alternative electronegativity difference method, we have;

\% \, Ionic \ Character = \left [18\times (\bigtriangleup E.N.)^{1.4}  \right ] \%

Where:

Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36

Therefore;

\% \, Ionic \ Character = \left [18\times (0.36)^{1.4}  \right ] \% = 4.3 \%

Hence the percentage ionic character = 4.3% ≈ 4%

the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%

The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;

The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.

The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.

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