Answer:
A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.
Explanation:
Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.
An example of a polysaccharide is starch, or glycogen.
Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.
Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.
Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.
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Answer:</h3>
19.3 g/cm³
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Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>