Answer: a. +2, cation and magnesium ion .
b. -1, anion, chloride
c. -2, anion, oxide
d. +1. cation , potassium ion
Explanation:
When an atom accepts an electron negative charge is created on atom and is called as anion.
When atom loses an electron positive charge is created on atom and is called as cation.
Magnesium (Mg) with atomic number of 12 has electronic configuration of 2,8,2 and thus it can lose 2 electrons to form 
cation and becomes magnesium ion.
Chlorine (Cl) with atomic number of 17 has electronic configuration of 2,8,7 and thus it can gain 1 electron to form 
anion and becomes chloride.
Oxygen (O) with atomic number of 8 has electronic configuration of 2,6 and thus it can gain 2 electrons to form 
anion and becomes oxide.
Potassium (K) with atomic number of 19 has electronic configuration of 2,8,8,1 and thus it can lose 1 electron to form 
cation and becomes potassium ion.
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
I believe it is the second option
Answer:
The correct answer is Option C (E1) and Option B (carbocation).
Explanation:
- Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
- Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.
Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.
Answer:
% yield = 62.21 %
Explanation:
- C2H4(g) + H2O(l) → C2H6O(l)
∴ mass C2H4(g) = 4.6 g
∴ mass C2H6O(l) = 4.7 g
- % yield = ((real yield)/(theoretical yield))×100
theretical yield:
∴ molar mass C2H4(g) = 28.05 g/mol
⇒ mol C2H4(g) = (4.6 g)*(mol/28.05 g) = 0.164 mol
⇒ mol C2H6O(l) = (0.164 mol C2H4)*(mol C2H6O(l)/molC2H4(g))
⇒ mol C2H6O(l) = 0.164 mol
∴ molar mass C2H6O(l) = 46.07 g/mol
⇒ mass C2H6O(l) = (0.164 mol)*(46.07 g/mol) = 7.55 g
⇒ theoretical yield = 7.55 g
⇒ % yield = (4.7 g)/(7.55 g))*100
⇒ % yield = 62.21 %
