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Debora [2.8K]
3 years ago
14

A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. S

olution A: 2.57 % (m/v) NaCl Solution B: 3.78 % (m/v) glucose Solution C: distilled H2O Solution D: 6.15 % (m/v) glucose Solution E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl
Chemistry
1 answer:
Burka [1]3 years ago
8 0

Answer:

Solution A: crenation

Solution B: hemolysis

Solution C: hemolysis

Solution D: crenation

Solution E: crenation

Explanation:

Hemolysis is the rupturing of red blood cells, which results in the release of hemoglobin (from within the red blood cells) into the plasma. If a red blood cell is placed in a hypotonic solution, water will flow into the cell, the cell will swell and hemolysis will.

Crenation: when a red blood cell is placed in a <em>hypertonic solution (</em>such as highly saline solution), the red blood cell will lose water(osmosis) and will shrink in size. The red blood cell has undergone crenation.

In order for a red blood cell to prevent from undergoing hemolysis or crenation, the cell must be placed in an<em> isotonic solution, </em>i.e either in <u>0.9% (m/v) NaCl solution</u> or <u>5% glucose solution</u>

  • Solution B and Solution C are hypotonic solution, thus red blood cell has undergone hemolysis.
  • Solution A, D and E are hypertonic solution. thus red blood cell has undergone crenation
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Which of the following represents the least number of molecues?
Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}

So, let´s get the number of molecules for each of the options.

a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules

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c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules

d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules

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Which of these formulas is NOT consistent with the guidelines for creating ionic compounds? Mark all that apply.
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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
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