Hope this helps :) I didn’t know how to write subscripts so I wrote it down on some paper.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
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Magnesium Peroxydisulphate with formula of MgS₂O₈.
<h3><u>Explanation:</u></h3>
The question clearly stands on the concept of molarity and atomic weights.
The atomic weight of magnesium = 24.
The atomic weight of sulphur = 32.
The atomic weight of oxygen = 16.
The amount of magnesium present = 0.248 g
The amount of sulphur present = 0.665 g.
The amount of oxygen present = 1.31g.
So, moles of magnesium present = 0.01 moles.
Moles of sulphur present = 0.02 moles.
Moles of oxygen present = 0.08 moles.
So, mole ratio of the compound as magnesium : sulphur : oxygen = 1:2:8.
So the compound is Magnesium Peroxydisulphate with formula of MgS₂O₈.