Answer:
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ is 4.11 × 10⁻³ grams
Explanation:
The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles
Therefore, we have;
One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂
3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂
The mass of one mole of Cl₂ = 70.906 g/mol
The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams
Therefore;
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ = 4.11 × 10⁻³ grams.
Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
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