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Rzqust [24]
3 years ago
11

You are investigating an organic compound. You discover that it is an aromatic compound often used as flavoring. What type of co

mpound is it likely to be?
Chemistry
2 answers:
LekaFEV [45]3 years ago
8 0

Answer:

Ester

Explanation: It is likely to be ester. Ester is a chemical compound that derives from an acid. It can be organic or inorganic. Esters are commonly used to add flavours. Each ester's flavor is different and most have different flavor thresholds. They are used in products suchs as beer, honey, rum.

Dafna1 [17]3 years ago
5 0
Aldehyde is the answer for Plato users!
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What is the bond angle in a tetrahedral molecule?
levacccp [35]
109.5

tetrahedral shape:
number of electron pair = 4,
number of bonded pair = 4,
number of lone pair = 0.
7 0
3 years ago
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A buffer consists of 0.45 M CH3COOH (acetic acid) and 0.35 M CH3COONa. The Ka of acetic acid is 1.8 x 10-5 . a) Calculate the pH
Zigmanuir [339]

Answer:

A) pH of Buffer solution = 4.59

B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original    buffer solution = 4.65

Explanation:

This  is the Henderson-Hasselbalch Equation:

 pH = pKa + log\frac{[conjugate base]}{[acid]}

to calculate the pH of the following Buffer solutions.

8 0
3 years ago
What does the thumping on the chest signal mean?
ankoles [38]
It means that you are the brother or friend of the person doing the thumping
3 0
3 years ago
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A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
3 years ago
Consider the equilibrium: HCOOH(aq) + F-(aq) <----> HCOO-(aq) + HF (aq) Given that the Ka of HCOOH = 1.8 x 10-4 and the Ka
Aleks [24]

Hey there!:

K = Ka * Kb / Kw

Ka = 1.8*10⁻⁴

Kb = 10⁻¹⁴ / 6.8*10⁻⁴

K =  1.8*10⁻⁴ * ( 10⁻¹⁴/ 6.8*10⁻⁴ ) * ( 1 / 10⁻¹⁴ )

K =  = 1.8 / 6.8

K = 0.265

Answer A

Therefore:

K is less than on the forward reaction is not favorable .


Hope That helps!

8 0
3 years ago
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