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Afina-wow [57]
3 years ago
7

Is the paramecium a prokaryote or a eukaryote

Chemistry
1 answer:
dimaraw [331]3 years ago
8 0
<span>The kingdom, protista’s characteristics are that the organism (not a plant, animal or fungus) are:<span> unicellular however some are multicellular like algae, are heterotrophic or autotrophic, others lives in water while some live in moist areas or human body, have a nucleus, cellular respiration is primarily aerobic, some are pathogenic (e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans.), Zooflagellates – flagellates (e.g. Trypanosoma gambiense), Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span></span>
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A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
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Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
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