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3 years ago

The bohr shift on the oxygen-hemoglobin dissociation curve is produced by changes in

Physics

1 answer:

Alisiya [41]3 years ago

7 0

Answer:

THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.

Explanation:

The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.

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2 years ago

A 0.250-kg aluminum bowl holding 0.800 kg of soup at 27.6°C is placed in a freezer. What is the final temperature if 424 kJ of e

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Answer:

16.32 °C

Explanation:

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Thus;

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Q' = 424 kJ - Q

Let's convert 424 KJ to Kcal

424 KJ = 424/4.184 Kcal = 101.3384 Kcal

Thus;

Q' = 101.3384 - 23.5635

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Q_f = Q' - mL

Where;

m = m_s = 0.8 kg

L = 79.8 kcal/kg

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Q_f = 77.7749 - (0.8 × 79.8)

Q_f = 13.9349 Kcal

Then final temperature will be;

T_f = Q_f/((m_b•c_b) + (m_s•c_s))

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2 years ago

A cannon ball is shot horizontally off a 37.0 m cliff and lands a distance of 18.5 m

4vir4ik [10]

Answer:

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Explanation:

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Now, in the vertical direction (coincident with the y- axis) , as both movements are independent each other, initial velocity is zero, so we can write the following equation for the vertical displacement:

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$t = \sqrt{\frac{2*\Delta h}{g} } =\sqrt{\frac{2*37.0m}{9.8m/s2}} = 2.75 s (3)$

Taking t₀ = 0, ⇒ Δt = t
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As the horizontal velocity is constant, the initial horizontal velocity is just the average one, i.e., 6.73 m/s.

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2 years ago