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sp2606 [1]
2 years ago
5

A blood specimen is collected in a heparin-containing tube for calcium and magnesium determination. Upon centrifugation, the pla

sma appeared hemolyzed. How would this affect the magnesium value?
Chemistry
1 answer:
Firdavs [7]2 years ago
3 0

Answer:

False hypermagnesaemia

Explanation:

Magnesium is predominantly an intracellular divalent action and it is important for optimal cellular function. It is an essential cofactor to many enzymes as well as being important for membrane function.

The body contain about 1 mol (approximately 25g) of magnesium.

Hemolyzed specimen will result in release of magnesium from intracellular to extracellular, causing elevation of the blood level of magnesium. This is a false hypermagnesaemia because it doesn't represent the real blood level.

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Because there no choices there, most probably it should be about hydrogen bonding in the water, which is the most important intermolecular force in water.
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3 years ago
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation
sleet_krkn [62]

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

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3 years ago
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Answer:

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3 years ago
The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

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3 years ago
What is the volume of 12.0 moles of Cl2 gas at STP?
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The answer has to be 22.4L
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2 years ago
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