Empirical formula = mol ratio
mol H = 9.42 x 10²⁴ : 6.02 x 10²³ = 15.648
mol N = 3/9 x 15.648 = 5.216
mass N = 5.216 x 14 g/mol =73.024 g
Bond Order = [Σ (bonding e-) - Σ (antibonding e-)]/2
<span>Be2 = 4e = σ1(2e) σ2*(2e) σ3(0) π1(0) π2*(0) σ4*(0) bo = 0 </span>
<span>[Be2]+ = 3e = σ1(2e) σ2*(1e) σ3(0) π1(0) π2*(0) σ4*(0) bo = 0.5 </span>
<span>[Be2]+ would be more likely to exist since it has a bond order of 0.5 whereas Be2 has zero bond order</span>
soap and water would be the best soak it in water for 20 miuntes then wash it
Answer:
NH₄⁺: 0.340M
NH₃: 0.277M
Explanation:
A buffer is the mixture of a strong acid with its conjugate base.
For the buffer of NH₃ / NH₄⁺, moles of each one are:
NH₄⁺: 0.287L × (0.310mol / L) = 0.0890 moles
NH₃: 0.287L × (0.310mol / L) = 0.0890 moles
The reaction of HNO₃ with NH₃ is:
HNO₃ + NH₃ → NH₄⁺ + NO₃⁻
Moles of 1.50mL of 6.00M HNO₃ are:
1.50x10⁻³L × (6.00mol / L) = 9x10⁻³ moles of HNO₃. These moles are moles produced of NO₃⁻ and consumed of NH₃. Thus moles after reaction are:
NH₄⁺: 0.0890 moles + 0.0090 moles = <em>0.0980moles</em>
NH₃: 0.0890 moles + 0.0090 moles = <em>0.0800moles</em>
As total volume is 287.00mL + 1.50mL = 288.50mL (0.28850L), concentrations are:
NH₄⁺: 0.0980moles / 0.28850L = 0.340M
NH₃: 0.0800moles / 0.28850L = 0.277M
<span>1.87 grams
Looking at the balanced equation, it indicates that for every mole of rust as a reactant, you need 6 moles of oxalic acid. So let's see how many moles of oxalic acid we have by multiplying the molarity of the solution by the quantity.
7.00x10^2 ml / 1000 ml/l * 0.100 M = 7.00x10^-2 mol
So the number of moles of rules that can be converted is
7.00x10^-2 mol / 6 = 1.17x10^-2 mol
Now determine the molar mass of Fe2O3.
Atomic weight iron = 55.845
Atomic weight oxygen = 15.999
Molar mass Fe2O3 = 2 * 55.845 + 3 * 15.999 = 159.687 g/mol
Mass Fe2O3 = 159.687 g/mol * 1.17x10^-2 mol = 1.87 g
So the amount of rust that can be removed is 1.87 grams.</span>
