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Stels [109]
3 years ago
15

Complete the sentences about heme. Some terms will not be used. The prosthetic group of hemoglobin and myoglobin is . The organi

c ring component of heme is . Under normal conditions, the central atom of heme is . In , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system. The central atom has bonds: to nitrogen atoms in the porphyrin, one to a residue, and one to oxygen.
Chemistry
1 answer:
nevsk [136]3 years ago
7 0

Explanation:

Haemoglobin consists of heme unit which is comprised of an <u>Fe^{2+}</u> 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 - <u>Heme</u>

The organic ring component of heme is - <u>Porphyrin</u>

Under normal conditions, the central atom of heme is - <u>Fe^{2+}</u>

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

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

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True or false water readily dissolves most polar compounds
morpeh [17]
The answer is false, I hope this helps.
3 0
4 years ago
Read 2 more answers
An ice-skating rink has tubes under its floor to freeze the water. salt water is cooled well below the freezing point of water a
alex41 [277]
Correct Answer: Option g: <span>adding salt to water lowers its freezing point

Reason:
Freezing point is a colligative property. When a non-volatile solution is present in solution, it's freezing point decreases. This is referred as depression in freezing point (</span>ΔTf<span>). Extent of lowering in freezing point is dependent on number of particles present in system. Mathematically it is expressed as:

</span>ΔTf = Kf X m
<span>
where, m = molality of solution
Kf = cryoscopic constant. 

Hence, a</span><span>dding salt to water lowers the freezing point of solution.</span> 
6 0
3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
MAVERICK [17]

<u>Answer:</u> The percent yield of water is 46.9 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For sulfuric acid:</u>

Given mass of sulfuric acid = 72.6 g    (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol

  • <u>For NaOH:</u>

Given mass of NaOH = 77 g      (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 12.5 g  (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%

Hence, the percent yield of water is 46.9 %

3 0
3 years ago
If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limi
Archy [21]

First you need to calculate the number of moles of aluminium and copper chloride.

number of moles = mass / molecular weight

moles of Al = 512 / 27 = 19 moles

moles of CuCl = 1147 / 99 = 11.6 moles

From the reaction you see that:

if        2 moles of Al will react with 3 moles of CuCl

then  19 moles of Al will react with X moles of CuCl

X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.

6 0
3 years ago
Treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid gives the compound C18H29BrO in quantitative yield. The
Cloud [144]

Answer:

2,4,6-tri-tert-butylcyclohexa-2,5-dienone

Explanation:

1. Information from the formulas

C₁₈H₃₀O ⟶ C₁₈H₂₉BrO

A Br has replaced an H.

2. Information from the reaction

These look like the conditions for an electrophilic aromatic substitution.

3. Possible mechanism (Fig. 1)

The product must be highly symmetrical, because there are so few NMR signals.

(i) The bromonium ion attacks at the para position, forming a resonance-stabilized carbocation intermediate.

(ii) A bromide ion attacks the H of the hydroxyl group to form 2,4,6-tri-tert-butylcyclohexa-2,5-dienone.

4. Confirmatory evidence (Fig. 2)

(a) Infrared  

1630 cm⁻¹:  C=O stretch

1655 cm⁻¹ : C=C stretch

(b)NMR

1.2 (9h, s):    the 4-tert-butyl group

1.3 (18H, s): the 2- and 6- tert- butyl groups

6.9 (2H, s):    the alkene H atoms

The ratio is 9:18:2.

3 0
3 years ago
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