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creativ13 [48]
2 years ago
12

11. Which of the following best explains why Co, gas is easily compressible but solid CO, (dry ice) is incompressible?​

Chemistry
1 answer:
telo118 [61]2 years ago
4 0

Answer:

The molecules of solid CO2 are much closer together than the molecules of CO2 gas.

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Answer:

C would be the right answer

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The element silver exists in nature as two isotopes: 107Ag has a mass of 106.9051 u, and 109Ag has a mass of 108.9048 u. The ave
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Answer:

107Ag has abundance of 51.7%

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Explanation: Please see attachment for explanation

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Which of the following group 7A elements is the most reactive ?
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Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl
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Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

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If an effusion experiment, it required 45 seconds for
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