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hjlf
3 years ago
15

Carbohydrates, more commonly known as sugars, are made up of carbon, oxygen, and hydrogen atoms. The smallest unit of a carbohyd

rate is a monosaccharide. Two monosaccharides make up a disaccharide, and many monosaccharides make up a polysaccharide. Disaccharides and polysaccharides can be hydrolyzed back into the individual monosaccharide units.Select the statement that is incorrect.a.) Complex sugars are carbohydrates.b.) All carbohydrates have the general formula Cn(H2O)n.c.) Simple sugars are carbohydrates.d.) Carbohydrates contain only carbon, oxygen, and hydrogen atoms.
Chemistry
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

The incorrect statement is option B and D.

Explanation:

Carbohydrates are more generally is known as the sugars that are composed of the carbon, hydrogen and oxygen atoms. Monosaccharides are the single unit of carbohydrates while disaccharides are the sugars that contain two molecules of the sugar and similarly saccharides with multiple sugars are called polysaccharides. The general chemical formula of the carbohydrates Cn(H2O)n, however the complex sugars does not follow this formula. Amino sugars are contains NH2 in their formula so carbohydrates not necessarily contains only H, O, AND C always.

Thus, the incorrect answer is - option B and D.

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Explanation:

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When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
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Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

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c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

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n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

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