<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
This was an answer i found on google from Nasa
First write the molecular equation with states:
(NH4)2S (aq) + 2AgNO3(aq) → Ag2S (s) + 2NH4NO3
Now write a full ionic equation by separating into ions all substances that dissociate: anything (s) (g) or (l) does not dissociate
2NH4 + (aq) + S 2-(aq) + 2Ag+ (aq) + 2NO3- (aq) → Ag2S(s) + 2NH4 + (aq) + 2NO3- (aq)
To write the NET IONIC equation, inspect the full ionic equation above and delete anything that appears on both sides of the → sign:
Net ionic equation:
S 2-(aq) + 2Ag + (aq) → Ag2S(s)