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.
D. the hyper-dimes
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True, because diatomic elements (H2, O2, F2, Br2, I2, N2, Cl2) consist of only one element but are molecules with covalent bonds.
Answer:
A.
Explanation:
Power is measured in Watts, and can we calculated by taking the work done in joules / the time in seconds. Thus, 1 Watt is 1 joule per 1 second (1 Watt = 1J/s)