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k0ka [10]
3 years ago
11

An animal cell placed in a hypertonic solution will shrink in a process called crenation. An animal cell placed in a hypotonic s

olution will swell and potentially burst in a process called hemolysis. To prevent crenation or hemolysis, an animal cell must be placed in an isotonic solution such as 0.9% (m/v) NaCl or 5.0% (m/v) glucose. This does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes. A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur.
Solution A: 3.21% (m/v) NaCl
Solution B: 1.65% (m/v) glucose
Solution C: distilled H2O
Solution D: 6.97% (m/v) glucose
Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl
Drag each solution to the appropriate bin.
Chemistry
1 answer:
tresset_1 [31]3 years ago
3 0

Answer:

Solution A: 3.21% (m/v) NaCl  ⇒ CRENATION

Solution B: 1.65% (m/v) glucose  ⇒ HEMOLYSIS

Solution C: distilled H₂O  ⇒ HEMOLYSIS

Solution D: 6.97% (m/v) glucose  ⇒ CRENATION

Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl ⇒ CRENATION

Explanation:

Isotonic solution = 0.9% (m/v) NaCl or 5.0% (m/v) glucose

<u>Crenation</u> will occur if the solution has a concentration higher than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypertonic solution). This will occur in the following solutions:

Solution A: 3.21% (m/v) NaCl  > 0.9% (m/v) NaCl

Solution D: 6.97% (m/v) glucose  > 5.0% (m/v) glucose

Solution E: 5.0% (m/v) glucose <u>and</u> 0.9%(m/v) NaCl (the addition of the 2 components exceeds the osmotic pressure permitted).

<u>Hemolysis</u> will occur if the solution has a concentration lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypotonic solution). This is the case of:

Solution B: 1.65% (m/v) glucose < 5.0% (m/v) glucose

Solution C: distilled H₂O = 0% glucose/0% NaCl < 0.9% (m/v) NaCl or 5.0% (m/v) glucose

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Answer:

Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

Explanation:

This is the reaction:

NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O

In 100 g of compound we have:

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Now we divide each  between the molar mass:

28.2 g / 14 g/mol = 2.01 mol

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And we divide again between the lowest value of moles

2.01 mol / 0.671 mol → 3

0.671 mol / 0.671 mol → 1

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2.68 mol / 0.671 mol → 4

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Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

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