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
Ionic compounds generally occur between metals and non-metals due to their large electronegativity difference. You can simple go down Group 1 and Group 17 of the periodic table.
Examples:
NaCl (Sodium Chloride)
KCl (Potassium Chloride)
RbCl (Rubidium Chloride)
CsCl (Cesium Chloride)
Explanation:
Ok so water is H2O and cabon dioxide in the air is CO2, so the water goes through the carbon dioxide and makes acid rain H2SO4.
Now we have our limestone which is CaCO3.
What happens is that the acid breaks apart our limestone into Ca2+ and CO3 2-. This then reforms into Calcium bicarbonate Ca(CO3)2.
Calcium bicarbonate is soluble in water and is hence washed away by the rain eroding the limestone.
Answer:
The pH decreases.
Explanation:
Hello,
In organic chemistry, oxidation accounts for either the increasing of C-O bonds or the increasing in the oxygen atoms into the molecule. Thus, if we consider the oxidation from benzyl alcohol to benzoic acid, there will be a carboxyl functional group instead of a hydroxile one. Now, the presence of the polar -O-H bonds that are ionizable, there will be a 
releasing causing the pH to decrease (increase acidity).
Regards.
Explanation:
First Reaction;
Ca + ZnCl2 --> CaCl2 + Zn
Oxidized Reactant: Ca. There is increase in oxidation number from 0 to +2
Reduced Reactant: Zn. There is decrease in oxidation number form +2 to 0
Second Reaction:
FeI2 + Mg --> Fe + MgI2
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Fe. There is decrease in oxidation number form +2 to 0
Third Reaction;
Mg + 2AgNO3 --> Mg(NO3)2 + Ag
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Ag. There is decrease in oxidation number form +1 to 0
