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nirvana33 [79]
3 years ago
6

Glucose (c6h12o6) has a single carbonyl group (-c=o) in its linear form. based on the number of oxygen atoms in glucose, how man

y hydroxyl groups (-oh) would you expect glucose to have?

Chemistry
1 answer:
sveta [45]3 years ago
3 0
Glucose is a hexose monosaccharide. It is one of the three major monosaccharides along with fructose and galactose. These are carbohydrates with a general formula of Cₓ(H₂O)ₓ, where x could be any number.

Now, you don't have to know the structural formula of the glucose to answer this. Just account all the elements in the glucose. You know that there are 6 oxygen atoms all in all. One of them belongs to the single carbonyl group. Consequently, that would mean that the remaining 5 oxygen atoms bond with hydrogen atoms to form 5 OH groups.

Just to be sure let us refer to the structural formula of glucose shown in the picture. It indeed has 5 OH groups.

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Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at <u><em>30 and 2204</em></u><em> </em>°C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

<em>Melting point</em> is the temperature at which the substace changes its state from<em> solid to liquid</em>. During that change, <em>the temperature</em> of the substance <em>does not change</em>, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means <em>that portion of the diagram is flat</em>.

The same is valid during<em> boiling</em>: the temperature remains constant while the substance is passing<em> from liquid to gas</em> at the boiling point.

Questoin 2:

The diagram for methane will have a <u><em>diagonal </em></u>line representing the <u><em>liquid phase</em></u> between -183°C and -162°C.

Between the <em>melting</em> (-183°C) and<em> boiling</em> (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a <em>diagonal</em> (slant) <em>line</em> between those points. During this interval the heat is used to <em>increase the temperature</em> and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at <u><em>2856</em></u><u> </u>°C of <u><em>the top horizontal line</em></u>

<u><em></em></u>

The table shows that the<em> boiling point</em> of gold is 2,856°C.

In a <em>temperature-vs.-time diagram</em> the<em> temperature is represented on the vertical axis (y-value)</em> and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during <em>boiling,</em> the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the <em>top horizontal line in the diagram</em>.

Question 4.

For nitrogen, the line at -210°C will be<u><em> flat</em></u>, which represents <u><em>the change from a solid to a liquid</em></u>

<u><em></em></u>

The table shows that the <em>melting point </em>of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

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