Answer:
We are heating the sample repeatedly to become a pure compound of only MgSO4 (withot H2O) and a constant mass.
Explanation:
Step 1: Data given
Mass of MgSO4·7H2O = 5.06 grams
The remaining MgSO4 had a constant mass of 2.47 grams.
Step 2: Explain why the sample in the crucible was heated repeatedly until the sample had a constant mass.
Before heating the compound has magnesium sulfate and water.
The total mass of this compound is 5.06 grams
By heating we try to eliminate the water.
After heating there remain mgSO4 with a mass of 2.47 grams
This means 5.06 - 2.47 = 2.59 grams is water. All of this is eliminated.
The heating process happens repeatedly to make sure the final compound is pure. So the 2.47 grams os only MgSO4. If the mass would not be constant. It means the compound is not pure, the not all the water is eliminated yet.
So we are heating the sample repeatedly to become a pure compound of only MgSO4 (withot H2O) and a constant mass.
Answer:
they just add heat, then it turns to water vapor
Explanation:
The first h20 is water the arrow represents adding heat, and the second h20 represent water vapor (water as a gas)
Two sublevels of the same principal energy level differ from each other through shape and size.
There are mainly 4 energy level s, p, d and f.
The s level has one orbital and one orbital have two electrons. So the maximum number of electron in s sublevel is 2.
The p level has three orbital and one orbital have two electrons. So the maximum number of electron in s sublevel is 6.
The d level has five orbital and one orbital have two electrons. So the maximum number of electron in s sublevel is 10.
The f level has 7 orbital and one orbital have two electrons. So the maximum number of electron in s sublevel is 14.
They may be differ in magnetic level.
Thus, we concluded that Two sublevels of the same principal energy level differ from each other through shape and size.
learn more about energy level:
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I don’t get it. What is the question asking?
Answer:
1.99 M
Explanation:
The molar mass of sodium thiosulfate (solute) is 158.11 g/mol. The moles corresponding to 110 grams are:
110 g × (1 mol/158.11 g) = 0.696 mol
The volume of solution is 350 mL = 0.350 L.
The molarity of sodium thiosulfate is:
M = moles of solute / liters of solution
M = 0.696 mol / 0.350 L
M = 1.99 M