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kodGreya [7K]
2 years ago
10

. Hemoglobin (Hb) and oxygen gas form a complex (HbO2) that carries oxygen throughout the human body. Unfortunately, carbon mono

xide also binds to hemoglobin so that an equilibrium is established. Carbon monoxide poisoning occurs when the concentration of HbO2 in the blood is reduced. HbO2 CO HbCO O2 The first aid for a person suffering from carbon monoxide poisoning is to (1) remove them to an area of fresh air, and (2) administer oxygen. Using the principles of equilibrium, explain how each of these helps to restore the HbO2 concentration.
Chemistry
1 answer:
sp2606 [1]2 years ago
5 0

Answer:

1. Removing them to an area of fresh air. This helps to prevents further poisoning by the carbon monoxide and increase the amount of oxygen entering into the body. This will help to reduce the concentration of carbon monoxide binding oxygen

2. Administering pure oxygen goes a long way to enhance ventilation and increase the oxygen saturation to 100%. This will help to overcome the effect of the carbon monoxide and promote more hemoglobin binding

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1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore
Alina [70]

Answer :

(1) The theoretical yield of product, MgO is, 1.257 grams.

(2) The percent yield of MgO is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.03142 moles of Mg react to give 0.03142 moles of MgO

Now we have to calculate the mass of MgO

\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO

\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g

Theoretical yield of MgO = 1.257 g

Experimental yield of MgO = 0.8922 g

Now we have to calculate the percent yield of MgO

\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100

\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%

Therefore, the percent yield of MgO is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0
2 years ago
Jose wants to know the effects of mass, force, and surface type on the motion of a small box. He has one spring, one box, and an
krok68 [10]

Answer:

He changed more than one variable at a time

Explanation:

6 0
2 years ago
Oran fills in the table below to organize information about the gas laws.
podryga [215]

Answer:2,4&5 A.

Explanation:

6 0
2 years ago
Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
JulijaS [17]

Answer : The correct option is, 30.9^oC

Explanation :

Formula used :

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released = 24 KJ

m = mass of bomb calorimeter = 1.30 Kg

c = specific heat = 3.41J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25.5^oC

Now put all the given values in the above formula, we get  the final temperature of the calorimeter.

q=m\times c\times (T_{final}-T_{initial})

24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC

T_{final}=30.9^oC

Therefore, the final temperature of the calorimeter is, 30.9^oC

5 0
2 years ago
Read 2 more answers
A beaker has 0.2 M of Na2SO4. What will be the concentration of sodium and sulfate ions?
netineya [11]

The concentration of sodium and sulphate ions are [Na^+] = 0.4 M, [SO_4^2-] = 0.2 M

Explanation:

The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.

In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.

Assuming 100% dissociation,

1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.

So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.

7 0
2 years ago
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