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True [87]
2 years ago
6

In the early twentieth century, what was Alexander Fleming’s hypothesis concerning the relationship between a yellow-green mold

and certain strains of bacteria?
a. He thought the bacteria had released a chemical that caused the mold to grow. b.He thought the bacteria fed on the mold, causing it to shrink and disappear.
c. He thought the mold had released a chemical that prevented the bacteria’s growth.
Chemistry
1 answer:
SpyIntel [72]2 years ago
4 0

Answer:

The correct answer is the option c)He thought the mold had released a chemical that prevented the bacteria’s growth.

Explanation:

In the 1920s, Alexander Fleming was working in his laboratory at St. Mary's Hospital in London when, almost by accident, he discovered a naturally growing substance that could attack certain bacteria. In one of his experiments, Fleming observed that colonies of a bacterium had been depleted or removed by a mold that grew on the same Petri dish. He observed that the bacteria furthest from the fungus had grown to produce large-sized colonies, while the colonies closest to the fungus were tiny. He determined that mold made a substance that could dissolve bacteria. The fungus was penicilium chrysogenum and thus Fleming called this substance penicillin, by the name of the mold that produces it. Thus, after several years of experiments in 1930, Howard Florey and Ernest Chain developed at Oxford University the procedures to produce pure penicillin from the fungus that Fleming isolated. Thus penicillin could be concentrated by Florey and Chain, and in 1945 they shared with Fleming the Nobel Prize in Medicine.

Then, <u><em>the correct answer is the option c)He thought the mold had released a chemical that prevented the bacteria’s growth.</em></u>

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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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