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SIZIF [17.4K]
3 years ago
10

Please help me and thank you

Chemistry
1 answer:
Anon25 [30]3 years ago
3 0
It is the first one because haven't you ever noticed when you hold a mirror a certain way the light can reflect a light up another surface? so, therefore, the answer that makes the most sense would be the first. :)
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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
Alex73 [517]

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

7 0
4 years ago
Which gas is used to take out blueprint​
tiny-mole [99]

The blueprint process

The best known is a process using ammonium ferric citrate and potassium . The paper is impregnated with a solution of ammonium ferric citr

7 0
3 years ago
You are working with a concentrated solution of ammonium hydroxide which place of safety equipment is most important to have on
natita [175]
I would say safety goggles
3 0
3 years ago
Read 2 more answers
The equilibrium constant for the reaction of fluorine gas with bromine gas at 300 K is 54.7 and the reaction is: Br2(g) + F2(g)
alisha [4.7K]

Answer:

The correct answer is 0.024 M

Explanation:

First we use an ICE table:

      Br₂(g)     +     F₂(g)    ⇔       2 BrF(g)

I      0.111 M          0.111 M                0

C      -x                   -x                      2 x

E      0.111 -x          0.111-x                2x

Then, we replace the concentrations of reactants and products in the Kc expression as follows:

Kc= \frac{[BrF ]^{2} }{[ F_{2} ][Br_{2}  ]}

Kc= \frac{(2x)^{2} }{(0.111-x)(0.111-x)}

54.7= \frac{4x^{2} }{(0.111-x)^{2} }

We can take the square root of each side of the equation and we obtain:

7.395= \frac{2x}{(0.111-x)}

0.111(7.395) - 7.395x= 2x

0.82 - 7.395x= 2x

0.82= 2x + 7.395x

⇒ x= 0.087

From the x value we can obtain the concentrations in the equilibrium:

[F₂]= [Br₂]= 0.111 -x= 0.111 - 0.087= 0.024 M

[BrF]= 2x= 2 x (0.087)= 0.174 M

So, the concentration of fluorine (F₂) at equilibrium is 0.024 M.

8 0
4 years ago
How many grams of NO will be produced from 60.0g of NO2 reacted with excess water in the following chemical reaction?
Lynna [10]

Answer:

Mass = 19.78 g

Explanation:

Given data:

Mass of NO produced = ?

Mass of NO₂ reacted = 60.0 g

Solution:

Chemical equation:

3NO₂(g) + H₂O(l)   →  2HNO₃(g) + NO(g)

Number of moles of NO₂:

Number of moles = mass/molar mass

Number of moles = 60.0 g/ 46 g/mol

Number of moles = 1.3 mol

Now we will compare the moles of NO₂ with NO.

                      NO₂         :           NO

                        3            :            1

                      1.3            :            1/3×1.3 = 0.43 mol

Mass of NO:

Mass = number of moles × molar mass

Mass = 0.43 mol × 46 g/mol

Mass = 19.78 g

6 0
4 years ago
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