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uysha [10]
3 years ago
9

Identify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues.

a. an increase in blood acidity near the tissues b. an increase in blood temperature near the tissues c. the presence of a pressure gradient for oxygen d. an exchange of ions in the erythrocytes
Chemistry
1 answer:
AlexFokin [52]3 years ago
7 0

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

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Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

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Or,

                              # of Particles  =  Moles × 6.022 × 10²³

So, from above relation it is found that 1 mole of any substance contains exactly 6.022 × 10²³ particles. Greater the number of moles greater will be the number of particles.
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