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MA_775_DIABLO [31]
3 years ago
5

Disaccharides are joined by glycosidic bonds formed between the anomeric carbon on one monosaccharide and a hydroxyl ( − OH ) gr

oup of another monosaccharide. Identify the types of linkages in each of the three disaccharides. Only place three of the linkages.
Chemistry
1 answer:
Anastaziya [24]3 years ago
7 0

<u>Solution and Explanation</u>

<u>1. </u>OH group is present above the plane in monosaccharide 1. Carbon atoms in both the rings are numbered in such way that carbon behind the oxygen at the right hand is C1 in the both rings. Carbon 1 bonds to carbon 2 of the next monosaccharide. Hence, the glycosidic linkage between disaccharide is  \beta 1 \rightarrow 2 .

2. OH group is present above the plane in monosaccharide 1 while in ring 2 it is present below the plane. Carbon atoms in both the rings are numbered in such way that carbon behind the oxygen at the right hand is C1 in the both rings. Carbon 2 bonds to carbon 1 of the next monosaccharide. Hence, the glycosidic linkage between disaccharide is   \beta 2 \rightarrow \alpha 1.

3. OH group is present below the plane in below the plane in both the rings. Carbon atoms in both the rings are numbered in such way that carbon behind the oxygen at the right hand is C1 in the both rings. Carbon 1 bonds to carbon 4 of the next monosaccharide. Hence, the glycosidic linkage between disaccharide is \alpha 1 \rightarrow 4.

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How many moles of potassium oxide (K2O) will be formed when 1.52 moles of potassium reacts with oxygen according to the followin
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How many grams of product can be produced by reacting 5.0 gram of aluminum and 22 grams of bromine? 2 Al + 3 Br2 &gt; 2 AlBr3
Crazy boy [7]

Answer:

The mass of AlBr3 is 24.5 grams

Explanation:

Step 1: Data given

Mass of aluminium = 5.0 grams

Mass of bromine = 22.0 grams

Molar mass of aluminium = 26.98 g/mol

Molar mass of br2= 159.8 g/mol

Step 2: The balancced equation

2 Al + 3 Br2 → 2 AlBr3

Step 3: Calculate moles Al

Moles Al = mass Al / molar mass Al

Moles Al = 5.0 grams / 26.98 g/mol

Moles Al = 0.185 moles

Step 4: Calculate moles Br

Moles Br = 22.0 grams / 159.8 g/mol

Moles Br = 0.138 moles

Step 5: Calculate limiting reactant

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)

Al is in excess. There will react 2/3*0.138 = 0.092 moles

There will remain 0.185- 0.092 = 0.093 moles Al

Step 6: Calculate moles AlBr3

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3

Step 7: Calculate mass AlBr3

Mass AlBr3 = moles * molar mass

Mass AlBr3 = 0.092 moles * 266.69 g/mol

Mass AlBr3 = 24.5 grams

The mass of AlBr3 is 24.5 grams

3 0
3 years ago
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