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.
the number of protons and the number of neutrons determine an element's mass number. :D
Answer:
–0.13 Pa.m²
Explanation:
From the question given above, the following data were obtained:
Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²
Measurement (Pa.m²) =?
We can convert from Pa.mm² to Pa.m² by doing the following:
1 Pa.mm² = 1×10¯⁶ Pa.m²
Therefore,
–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²
–1.3×10⁵ Pa.mm² = –0.13 Pa.m²
Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².
688x
Explanation- Your welcome