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uysha [10]
2 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]2 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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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
1 year ago
5. A guava with a mass of 0.200 kg has a weight of
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2 years ago
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All amino acids that are found in proteins, except for proline, contain a(n); Group of answer choices amino group carbonyl group
Jobisdone [24]

Answer:

amino group

Explanation:

There are twenty (20) amino acids in nature. Generally, each amino acid is structurally made up of a central carbon atom called alpha carbon attached to a hydrogen, carboxylic acid group (-COOH) and an amine group (-NH2). However, one particular amino acid called PROLINE posseses an exception to this.

Proline, which is the only cyclic amino acid, is also the only amino acid that forms a secondary amine group i.e. loss of hydrogen atoms in its amine group when in a protein structure. This means that when in a protein, PROLINE does not have an AMINE GROUP.

7 0
2 years ago
Based on the activity series provided, which reactants will form products?
allsm [11]

Answer: CuI₂ + Br₂

Explanation:

1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).

The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.

2) Choice 1: CuI₂ + Br₂

Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:

CuI₂ + Br₂ → CuBr₂ + I₂

Being I less active than Br, it cannot displace Br in CuBr₂.

3) Choice 2: Cl₂ + AlF₃

Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.

4) Choice 3: Br₂ + NaCl

Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.

5) Choice 4: CuF₂ + I₂

Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.

6 0
2 years ago
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Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to
Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode (m_{ch}) is given as :

\frac{m_{ch}}{M} ZF = It

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; (m_{ch}) = (Ad) \rho ; replacing that into above equation; we have:

\frac{(Ad) \rho}{M} ZF = It  ---- equation (1)

where;

A = area

d = thickness

\rho = density

From the above equation (1); The time required for coating can be calculated as;

[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons }  ] = (20 A) t

t = \frac{2100}{20}

= 105 s

8 0
2 years ago
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