Gold is the best pure substance. The answer is gold
Answer:
<h2>0.52 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
volume = final volume of water - initial volume of water
volume = 35 - 8 = 27 mL
We have
We have the final answer as
<h3>0.52 g/mL</h3>
Hope this helps you
Answer:
V₂ = 0.62 L
Explanation:
Given data:
Initial volume = 2.4 L
Initial temperature = 25°C
Final temperature = -196°C
Final volume = ?
Solution:
Initial temperature = 25°C (25+273 = 298 K)
Final temperature = -196°C ( -196+273 = 77 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 2.4 L × 77 K / 298 k
V₂ = 184.8 L.K / 298 K
V₂ = 0.62 L
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
Answer:
a. neutral
b. salts
c. salt
Explanation:
Organic salts are a dense number of ionic compounds with innumerable characteristics. They are previously derived from an organic compound, which has undergone a transformation that allows it to be a carrier of a charge, and that in addition, its chemical identity depends on the associated ion.
Organic salts are usually stronger acids or bases than inorganic salts. This is because, for example, in the amine salts, it has a positive charge due to its bond with an additional hydrogen: A + -H. Then, in contact with a base, donate the proton to be a neutral compound again
RA + H + B => RA + HB
H belongs to A, but it is written as it is involved in the neutralization reaction.
On the other hand, RA + can be a large molecule, unable to form solids with a crystalline network stable enough with the hydroxyl anion or oxyhydrile OH–.
When this is so, salt RA + OH– behaves as a strong base; even as basic as NaOH or KOH
