What is a material that does not allow heat to flow through easily?
1 answer:
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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:
Explanation:
Hello there!
In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.
(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:
And the moles of hydrogen due to the produced 1.50 grams of water:
Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:
Thus, the subscripts in the empirical formula are:
As they cannot be fractions.
(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:
Which are equal to the moles of the acid:
And the molar mass:
(3) Finally, since the molar mass of the empirical formula is:
7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol
Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:
Best regards!