Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: 
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29)
; Silver hydroxide at higher temperatures decomposes into black silver oxide: ![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29)
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate: 
. The latter precipitate is yellow.
The thing you MUST do FIRST is look for any H's, O's, or F's in the equation
1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA
2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)
ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet
for example::::
N2O4
the oxidation number of O is -2.
since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4
N has an oxidation number of +4.
more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so
oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion
Answer:
324.24 kPa
Explanation:
Given that;
Initial pressure P1 = 101325 Pa
V1= 400 ml
P2 = ?
V2= 125mL
From Boyle's law;
P1V1 = P2V2
P2 = P1V1/V2
P2= 101325 × 400/125
P2= 324.24 kPa
There are two oxygens on the left side of the equation and three on the right
The Protons determine the identity.
