I assume about 5 atoms are in one molecule of this product
B) 40%
The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.
Answer:
if i remember correctly it's B
Explanation:
Answer:
The minimum concentration of Cl⁻ that produces precipitation is 12.6M
Explanation:
The Ksp of PbCl₂ is expressed as:
PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
The Ksp is:
Ksp = 1.6 = [Pb²⁺] [Cl⁻]²
When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.
A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:
1.6 = [0.010M] [Cl⁻]²
160 = [Cl⁻]²
12.6M = [Cl⁻]
<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>
Answer:
0.404M
Explanation:
...<em>To make exactly 100.0mL of solution...</em>
Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.
To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:
<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>
3.26g * (1mol / 241.86g) =
0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =
<em>0.0404 moles of NO₃⁻</em>
In 100mL = 0.1L, the molar concentration is:
0.0404 moles of NO₃⁻ / 0.100L =
<h3>0.404M</h3>
