What is the mass of helium atom whose atomic weight is 4.003 g/mol?
1 answer:
Answer:
Atoms He (Avogadro’s number) → Moles of He (molar mass of He) → Mass of He
• molar mass of He (from the periodic table) = 4.003 g/mol
• Avogadro’s Number: Avogadro’s number gives us the number of entities present in 1 mole: 6.022 × 1023 He atoms in 1 mole of He
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Answer:
Trial Number of moles
1 0.001249mol
2 0.001232mol
3 0.001187 mol
Explanation:
To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.
Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.
<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.
In this case the solute is <em>NaOH</em>.
The formula is:
Solve for the <em>number of moles:</em>
Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.
Trial mL liters Number of moles
1 12.49 0.01249 0.01249liters × 0.1000M = 0.001249mol
2 12.32 0.01232 0.01232liters × 0.1000M = 0.001232mol
3 11.87 0.01187 0.01187liters × 0.1000M = 0.001187 mol
Answer: the empirical formula is C7H13O10N
Explanation:Please see attachment for explanation
