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ella [17]
3 years ago
5

Which of the following are bound to hemoglobin when hemoglobin is in the r-state? choose all that apply.

Chemistry
1 answer:
Mademuasel [1]3 years ago
7 0
Even though there is no followings, I will try to include in this answer all the possible answers. I am sure about two elements that bound to hemoglobin when hemoglobin is in the R-state :Fe+2 and O2. As you should know, <span> R-state of hemoglobin is a relaxed form that is also called</span> <span>oxyhemoglobin, so the O2 definitely must be mentioned in there.</span>
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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
Fed [463]

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

7 0
3 years ago
For which characteristic would a human MOST LIKELY selectively breed a house cat?
madreJ [45]
B. Coat color I’m pretty sure
8 0
3 years ago
Read 2 more answers
Phosphorous pentchloride (pcl5 ) is used as a chlorinating reagent in chlorination of organic compounds. what type of bonding oc
schepotkina [342]

<span>The type of bond that a Phosphorous pentachloride have is an Ionic Bonding. It is a form of chemical bond that encompasses the electrostatic attraction between oppositely charged ions which serves as the primary interaction happening in ionic compound. Phosphorus has 5 valence electrons and Chlorine has 7 valence electrons. Phosphorus contributes 1 electron to each chlorine and all the 6 achieve 8 electrons in the outer shell thus creating an ionic bond.</span>

5 0
3 years ago
Read 2 more answers
25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?
MAVERICK [17]

The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.

<h3>What is the molarity of 5% NaOH?</h3>

The molarity of 5% NaOH is 1.32 M

25 cc of NaOH neutralized 30cc of H₂SO₄ solution.

Equation of reaction is given below:

  • 2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O

Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M

  • Normality = Molarity × moles of H⁺ ions per mole of acid

moles of H⁺ ions per mole of H₂SO₄ = 2

Normality of H₂SO₄ = 0.55 x 2 = 1.1 N

In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.

Learn more about normality at: brainly.com/question/22817773

#SPJ1

7 0
1 year ago
A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.
Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

  • Concentration of the weak acid (Ca): 0.187 M
  • pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}

3 0
3 years ago
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