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 correct answer is q < 0. Hope this helps you!!
The molar solubility is 7.4×
M and the solubility is 7.4× g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→ 
+ 
S
- S S
Ksp = [ ] [ ] = [S] [ S ] = 
S = √ Ksp = √ 5. 5×
= 7.4×
The solubility =7.4× g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of is dissociates and form one mole of each and ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4× mol/L = 7.4× M
To learn more about molar solubility ,
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Answer:
The answer to your question is pH = 9.2; [OH] = 1.58 x 10⁻⁵
Explanation:
Data
[H⁺] = 6.0 x 10⁻¹⁰ M
pH = ?
pH definition. pH measures the hydrogen concentration. pH measures the acidity and the alkalinity of a solution. A solution is acid if the pH goes from 0 to 6.9, a neutral solution has a pH of 7 and an alkali solution has a pH from 7.1 to 14.
pOH measures the [OH⁻] concentration
Formula
pH = -log[H⁺]
pH + pOH = 14
-Substitution
pH = -log[6 x 10⁻¹⁰]
-Simplification
pH = - (-9.2)
pH = 9.2
-Calculate pOH
pOH = 14 - pH
pOH = 14 - 9.2
pOH = 4.8
-Calculate the concentration of OH⁻
pOH = -log[OH⁻]
[OH⁻] = -antilog4.8
[OH] = 1.58 x 10⁻⁵
The answer is, 699g
The idea here is that you need to use the mole ratio<span> tha exists between </span>ferric oxide<span>, </span><span><span>Fe2</span><span>O3</span></span><span>, and iron metal, </span>Fe<span>, to determine how many moles of the latter will be produced when </span>all the given mass<span> of the ferric oxide reacts...
</span>
