Gold alloys consist of most commonly Nickel!
Most commonly used in things like jewelry so it doesn't have such a heavy weight.
Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = ![\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
Answer:
b. Mg (s) + 2HCl (aq) --> MgCl2 (aq) + H2
Explanation:
A subscript applies only to the element it is attached to while coefficients apply to all elements in a compound.
"a" has 1 atom of each element on the left but 2 Cl on the right.
"c" has 2 Mg, 1 H, & 1 Cl on the left but 1 Mg, 2 H, and 2 Cl on the right.
"d" has 2 atoms of each element on the left but only 1 Mg on the right.
"b" is the only answer with equal numbers of atoms on the right and left sides of the equation.
Answer:
The correct answer is - (4) 1s2 2s2 2p5 3s2
Explanation:
An excited state is a state when the valence electron has moved to some other higher energy orbital, from its ground state orbital. The ground state has a lower energy level or sublevel. In this case, the higher energy level orbit fills before the lower energy level.
In option 4, the last electron is filled in higher energy orbit 3s2 before filling the lower or ground energy level 2p5, in the ground state it would be 1s2 2s2 2p6 3s1 instead of 1s2 2s2 2p5 3s2.
Thus, the correct answer is option 4.
