<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.
In chemical reactions, the actual yield is not the same as the expected yield . Actual yield is lower than the theoretical yield . Then we have to find the yield percentage. To see what percentage of the theoretical yield is the actual yield.
Percent yield = actual yield / theoretical yield x 100%
Percent yield = 24.6/55.9 x100%
Percent yield = 44%
Doubling the amplitude of a wave means that you increase the energy of a wave factor of quadrupling
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital.