Elements in this family have 1 valence electron?

Haemoglobin consists of heme unit which is comprised of an $Fe^{2+}$ and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.

The prosthetic group of hemoglobin and myoglobin is - Heme

The organic ring component of heme is - Porphyrin

Under normal conditions, the central atom of heme is - $Fe^{2+}$

In deoxyhemoglobin , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.

The central atom has six bonds: four to nitrogen atoms in the porphyrin, one to a histidine residue, and one to oxygen.

7 0

3 years ago

Identify the base-conjugate acid pair in this balanced equation:

Elan Coil [88]

<h3>Answer;</h3>

NH3/NH4+

<h3>Explanation;</h3>

From the equation;

NH3(aq)+HNO3(aq)→NH4+(aq)+NO3−(aq)

NH3 is the base; while NH4+ is the conjugate acid

HNO3 is the acid; while NO3- is the conjugate base

The conjugate base of a Brønsted-Lowry acid is species that is formed after an acid donates a proton while the conjugate acid of a Brønsted-Lowry base is the species formed after a base accepts a proton.

3 0

3 years ago

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A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin

gtnhenbr [62]

Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)

The ksp is defined as:

ksp = [borate²⁻] [Na⁺]²

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl× $\frac{1molB_{4}O_{7}^{2-}}{2molHCl}$ = 3,00x10⁻³ mol of B₄O₇²⁻

For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺

The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M

And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

ksp = 0,176

I hope it helps!

8 0

3 years ago