Although multiple values are given, our focus is on HCl.
<span>We have 215 mL (0.215 L) of 0.300 M HCl fully consumed in the reaction. It's important to recall that the number of moles is found by multiplying volume by molarity:</span>
moles = 0.215 L × 0.300 M
<span>moles = 0.0645 moles of HCl</span>
Response:
Clarification:
Hello,
In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:
Best wishes.
M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL
From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol:
3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10
Next convert moles to molecules using Avogadro’s number:
0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10
Each butane molecule contains 4 carbon atoms, so:
3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.
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