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3 years ago

Is cannabinol soluble in 1-octanol? Why or why not?

Chemistry

1 answer:

Vilka [71]3 years ago

3 0

Answer: Uhmmmm this is what i got...-

Explanation:

It is a matter of fact that the hydrophobic (water-hating) oily compounds of cannabinoids such as CBD, THC and others are not water soluble. The term "water soluble" refers to materials which dissolve in water in a homogenous manner by becoming molecules or ions (such as sugar, alcohol, and salt).

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Identify each type of titration curve. note that the analyte is stated first, followed by the titrant.

Ghella [55]

Each of the type of titration curve are as follows: Strong acid-strong base= starts small than increasesWeak acid-strong base= low then high (small difference) Weak base- strong acid= high to low. Have in mind that titraton curves generally contain the volume of the titrant as the independent variable and the ph of the solution as the dependent variable.

8 0

3 years ago

Is cake an acid base reaction

levacccp [35]

Answer:

What do you mean by this question, did you mean Coke, or what did you mean because I do not understand.

Explanation:

5 0

3 years ago

The speed of a car is 16 meters per second. What is this speed in miles per hour? What is

Dovator [93]

Conversion Factors:

1.61 km = 1 mile

60 sec = 1 min

60 min = 1 hour

Use dimensional analysis to solve.

$=\frac{16\ meters}{1\ second}$ × $\frac{1\ km}{1000\ meters}$ × $\frac{1 mile}{1.61\ km}$ × $\frac{60\ seconds}{1\ min}$ × $\frac{60\ min}{1\ hour}$

$=35.77639752$

= 36 miles/hour

If done properly, you should be left with the units miles/hour

3 0

2 years ago

All + HgCl, --------> AICI,

Free_Kalibri [48]

Answer:

64.7g

Explanation:

The balanced chemical equation of this question is as follows;

AlI + HgCl2 → HgI + AlCl2

Based on the above equation, 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide).

Using mole = mass/molar mass to convert mass of HgI to moles.

Molar mass of HgI = 200.59 + 127

= 327.59g/mol

Mole = 138/327.59

= 0.42mol

- If 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide)

- Then 0.42 mol of HgI will be produced by 0.42mol of AlI.

Using mole = mass/molar mass

Mass = mole × molar mass

Molar mass of AlI = 27 + 127

= 154g/mol

Mass of AlI = 0.42 × 154

= 64.7g of AlI

5 0

3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago