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sp2606 [1]
3 years ago
15

Amylose is a form of starch which has ________. only β-1,4-bonds between glucose units only α-1,4-links bonds glucose units hemi

acetal links joining glucose units carbon-carbon bonds joining glucose units both α-1,4-and β-1,4-bonds between glucose units g
Chemistry
1 answer:
forsale [732]3 years ago
5 0

Answer: Amylose is a form of starch which has only α-1,4-links bonds glucose units.

Explanation:

Amylose is a polysaccharide made up of α(1-4) bound glucose molecules. The carbon atoms on glucose are numbered, starting at the aldehyde (C=O) carbon, so, in amylose, the 1-carbon on one glucose molecule is linked to the 4-carbon on the next glucose molecule.

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What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare 100.0 milliliters of a 0.50 M acetic acid
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We are given with a 2.5 M stock solution of acetic acid and we are required to calculate the volume of the solution needed to prepare 100 milliliters of 0.5 M acetic acid solution. To solve this, we acquire the formula <span>Mconcentrated*Vconcentrated = Mdilute*Vdilute. That is 2.5 M*x=0.5M*100 ml where x is the volume of 2.5 M needed. x is equal to 20 ml. So we need 20 ml of 2.5 M solution and dilute to 100 ml using water as diluent.</span>
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3 years ago
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Answer:

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Explanation:

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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
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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.

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3 years ago
Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class de
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Explanation:

Rudolf Ludwig Carl Virchowas known as the founder of social medicine and also the father of modern pathology.

Virchow posited that all cells are gotten from already existing cells and he used this in his work towards cellular pathology, as it was made clear that diseases takes place at the cellular level. He posited that the cells that are malfunctioning cause diseases.

Based on the above analysis, the image that Tumu would most likely use in his assignment to feature Rudolf Virchow is the way root cells reproduce to increase root length.

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