Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
Answer : The correct answer is, (c) the number of neutrons
Explanation :
Isotope : It is defined as the element that have the same number of protons but have the different number of neutrons of each of the atom.
Atomic number is defined as the number of protons or number of electrons.
Atomic number = number of protons = number of electrons
Mass number is defined as the sum of number of protons and number of neutrons.
Number of neutrons = Mass number - Atomic number
For example : For Carbon - 13 isotope.
Mass number = 13
Atomic number = 6
Number of neutrons = Mass number - Atomic number
Number of neutrons = 13 - 6 = 7
Hence, the difference between the mass number of an isotope and its atomic number is the number of neutrons.
Answer: The answer is B The kinetic energy breaks the bonds between atoms.
Explanation:
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
The chemical equation that shows the reaction between nh3 and cuh206 is detailed as: [Cu(H2O)6]2+ (aq) + 2NH3(aq). —> [Cu(OH)2(H2O)4](s) + 2NH4 + (aq). the blue precipitate is Cu(OH)2(H2O)4 in which the blue color is caused by the Cu present in the solid.