0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Answer:
Gold is a metal, more specifically a transition metal, whereas Oxygen is a nonmetal, more specifically a reactive nonmetal. Using this information, you can compare and contrast metals, nonmetals, and metalloids.
Metals are:
Shiny
High melting point
Mostly silver or gray in color
Mostly solids at room temperature – Mercury (Hg) is a liquid at room temperature
Malleable – able to be hammered into a thin sheet
Ductile – able to be drawn/pulled into a wire
Good conductors of heat and electricity
Nonmetals are:
Dull
Low melting point
Brittle – break easily
Not malleable
Not ductile
Poor conductors of heat and electricity
Metalloids are:
Found on the “zig-zag” line on the Periodic Table of Elements
Have properties of both metals and nonmetals
Can be shiny or dull
Semiconductors – able to conduct electricity under certain conditions
Explanation:
Reccomend this site for questions llike these: https://ptable.com/#Properties
Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
