The equation of the reaction before balancing is
a0NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
PO₄³⁻ ion is balanced.
on the left side, theres 1 (NH₄⁺) ion and right side 3 (NH₄⁺) ions. Therefore if we put the coefficient for NH₄Cl, we will obtain the following equation
3 NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
3 Ag⁺ ions on the left side and 1 Ag⁺ ion on the right side, so if we put the coefficient of AgCl as 3, following equation obtained
3 NH₄Cl + a1Ag₃PO₄ → 3 AgCl + a3(NH₄)₃PO₄
Cl⁻ ions are also balanced now, 3 on either side.
a1 and a3 are 1 as those compounds are as it is, so coefficient is 1 for both
balanced equation is as follows
3 NH₄Cl + Ag₃PO₄ → 3 AgCl + (NH₄)₃PO₄
coefficients are
a0 - 3
a1 - 1
a2 - 3
a3 - 1
It will probably turned into a gumy
Becoing itsodim fromate t will have pb of 67
Answer:
This is a chemical symbol for Potassium Bromide
Answer:
The concentration of chloride ions in the final solution is 3 M.
Explanation:
The number of moles present in a solution can be calculated as follows:
number of moles = concentration in molarity * volume
In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻
For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:
number of moles of Cl⁻ = 2 * number of moles of CaCl₂
number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻
The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.
Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:
Concentration = number of moles of Cl⁻ / volume
Concentration = 150.2 mol / 50.1 l = 3.0 M
