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marin [14]
2 years ago
14

Draw the structure of beeswax.beeswax is made from the esterfication of a saturated 16-carbon fatty acid and a 30 carbon straigh

t chain primary alcohol.

Chemistry
1 answer:
Cloud [144]2 years ago
3 0

Answer:

Triacontyl palmitate

Explanation:

In this case, we have a reaction between an acid and an alcohol. When we put together these kind of compounds an ester is produced. This reaction is called <u>"esterification"</u>.

In our case, the alcohol is a structure with 30 carbon in which the "OH" group is bonded on carbon 1. The name of this compound is <u>"n-triacontanol"</u>. The acid is a structure in which we have 16 carbon in which the "COOH" group is placed on carbon 1. The name of this compound is <u>"palmitic acid"</u>. The ester produced by the acid and the alcohol is <u>"Triacontyl palmitate"</u>.

See figure 1.

I hope it helps!

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Acetone major species present when dissolved in water
jek_recluse [69]

Answer: acetone molecule ( CH₃-CO-CH₃)


Explanation:


1) Acetone is CH₃-CO-CH₃


2) That is a molecule (build up of covalent bonds).


3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.


This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.


That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).



3 0
2 years ago
Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na
Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}

valence according to the amount of salt anion  

Input the value :

\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3

7 0
2 years ago
What is the freezing point of a 1.40 m aqueous solution of a nonvolatile un-ionized solute? (the freezing point depression const
Alina [70]
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m

where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m

</span>∴ ΔTf = Kf X m  = 1.86 X 1.40 = 2.604 oC

Now, for water freezing point = 0 oC

∴Freezing point of solution = -2.604 oC
6 0
2 years ago
Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [
serious [3.7K]

Answer: The rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

Explanation:

The given chemical reaction is:

2NO_2\rightarrow 2NO+O_2

The rate of the reaction for disappearance of NO_2 is given as:

\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}

Or,

\text{Rate of disappearance of }NO_2=-\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of NO_2 = 0.00650 M

C_1 = initial concentration of NO_2 = 0.0100 M

t_2 = final time = 100 minutes

t_1 = initial time = 0 minutes

Putting values in above equation, we get:

\text{Rate of disappearance of }NO_2=-\frac{0.00650-0.0100}{100-0}\\\\\text{Rate of disappearance of }NO_2=3.5\times 10^{-5}M/s

Hence, the rate of disappearance of NO_2 is 3.5\times 10^{-5}M/s

6 0
2 years ago
Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount
shtirl [24]
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

5 0
2 years ago
Read 2 more answers
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