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EastWind [94]
3 years ago
12

Although cannabinol and methyl alcohol are both alcohols, cannabinol is very slightly soluble in methyl alcohol at room temperat

ure. Explain.
Chemistry
2 answers:
xxMikexx [17]3 years ago
8 0

Cannabinol is a psychoactive cannabinoid that is found in trace amounts in Cannabis. This is most often found in aged Cannabis. Oxidized cannabis products and traditionally produced hashish are high in cannabinol. Cannabinol is slightly soluble in methyl alcohol at room temperature. This is because it is not very polar (two non-polar rings and one polar ring) and is solid at room temperature.

olganol [36]3 years ago
5 0

Hey there!:

It is a solid at room temperature.

Take a look at the structure of the compound. It has 3 rings total, two of which are phenyl rings, 3 methyl groups and a pentyl group while there is only one OH. That is a lot of non-polar groups for just one polar group. This is why it is sparingly soluble in MeOH at RT, it really is not a very polar molecule (the general rule of thumb is 5 C's for every polar group for the molecular to be considered polar, this of course is a generalization).

I hope that helps.

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1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a
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Answer:

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃  +   Al  →  3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

                               Al           :            Ag

                                1            :             3

                                0.223   :         3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂  →   2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

                         HgO         :         Hg

                            2            :          2

                          3.12          :       3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂  + 5O₂   →  2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

                 N₂O₅          :           O₂

                     2            :             5

                    12.99      :         5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆  + 15O₂   →  12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

                  C₆H₆         :           CO₂

                     2            :             12

                    0.103      :         12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

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Calculate the volume which 1.00 mole of a gas occupies at 1 atm and 298K?
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Answer:

25.45 Liters

Explanation:

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (1mole)(0.08206Latm/molK)(298K)/(1atm) = 25.45 Liters

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