Which of the following are base formulas? (Choose 2)

D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match th

Aleks04 [339]

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

6 0

3 years ago

Which change would increase the pressure exerted by a gas?

Gelneren [198K]

Answer:

C - raising the temperature of a gas

Explanation:

as you raise temperature, kinetic energy rises, and so does pressure

7 0

3 years ago

For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this

Salsk061 [2.6K]

The following are the steps to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3 N = +5

O = -2 O = -2

Ag = +1 Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

NO2– -----> NO3–

Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

NO2– + H2O -----> NO3–

Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

NO2– + H2O -----> NO3– + 2H+

Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

NO2– + H2O -----> NO3– + 2H+ + 2e-

Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

NO2– + H2O -----> NO3– + 2H+ + 2e-

2 Ag+ + 2e------> 2 Ag(s)

3 0

3 years ago

Which one of the following best represents the predicted approximate chemical shift and coupling for the hydrogen(s) indicated w

melomori [17]

Answer:

The answer is quartet 2.40 ppm.

Note: Kindly find an attached image below for the part of the solution to this question

Sources: The image was researched from Course hero platform

Explanation:

Solution

Multiplicity or (n+1) rule:

It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.

Multiplicity =(n+1)

Thus

The non equivalent protons which are attached from adjacent atom is denoted by n.

Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:

The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:

Non- equivalent = 3

Multiplicity = (3 +1)

= 4

= Quartet for 2H

A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.

Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.

5 0

3 years ago

Calculate moles of oxygen atoms in 0.68mol of KMnO4

Dominik [7]

Alright, so that means we have 0.68 mol of the compound

For each 1 mol of the compound, we have 4*1 oxygens (because there are four oxygens in the formula)
Therefore for each 0.68 mol of the compound, we have 4*0.68 moles of oxygen!

6 0

3 years ago