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mojhsa [17]
2 years ago
11

You inoculate a bacterium into three tubes of nutrient broth medium containing added salt: 0.5% NaCl, 5% NaCl, and 15% NaCl. Aft

er incubation, you notice significant turbidity in the 15% NaCl medium and modest turbidity in the 5% NaCl medium. There is no growth in the third tube. This organism would be termed ________.
Chemistry
1 answer:
g100num [7]2 years ago
5 0

Answer:

Halophile.

Explanation:

Halophile microorganisms are microorganisms that  require very large amounts of NaCl. If the concentration of NaCl is very little, there will be no growth. Ih this case, given that it grows between 5% and 15%, it can be considered a moderate halophile.

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so
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Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result  in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl
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