Answer:
if i remember correctly i beleive its A 1.8 x 10^24
but im not for sure also i think you forgot the 24
Explanation:
Answer:
semiconducting,tellurium
Explanation:
just completed the assignment
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
Answer:
C. Lose three electrons to have a full outer shell
Explanation:
Al is in Group 13 of the Periodic Table, so it has three valence electrons.
It must either lose three electrons or gain five to achieve a stable octet.
It is easier to lose three electrons than it is to gain five, so Al loses three electrons.
D. is wrong, for the same reason.
A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.
C. is wrong. Al is a metal, so it will lose electrons in a reaction.
Answer: i beleive it is fixation in edge 2020
Explanation:
