Answer:
The formula for velocity is v = Δs/Δt
Explanation:
Velocity equals change in speed divided by change in time.
Answer: ₉₈²⁵³Cf
253 is a superscript to the left of the symbol, Cf, which represents the mass number, and 98 is a subscript to the left of the same symbol, which represents the atomic number.
Explanation:
1) The alpha decay equation shows that the isotope Fm - 257, whose nucleus has 100 protons and 157 neutrons, emitted an alpha particle (a nucleus with 2 protons and 2 neutrons).
2) Therefore:
i) the mass number decreased in 4, from 257 to 257 - 4 = 253.
2) the atomic number decreased in 2, from 100 to 100 - 2 = 98.
3) Hence the formed atom has atomic number 98, which is californium, Cf, and the isotope is californium - 253.
4) The item that completes the given alpha decay reaction is:
₉₈²⁵³ Cf.
5) The complete alfpha decay reaction is:
₁₀₀²⁵⁷ Fm → ₉₈²⁵³Cf + ₂⁴He
You can verify the mass balance:
257 = 253 + 4, and
100 = 98 + 2
<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
I guess is weak alkaline. when the substance is more acidic, there will be less alkalinity
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!
