15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
Answer:
The isotopic mass of 41K is 40.9574 amu
Explanation:
Step 1: Data given
The isotopes are:
39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%
40K with an isotopic mass of 39.963999u
41K wit natural abundance of 6.7302 %
Average atomic mass =39.098 amu
Step 2: Calculate natural abundance of 40 K
100 % - 93.2581 % - 6.7302 %
100 % = 0.0117 %
Step 3: Calculate isotopic mass of 41K
39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302
39.098 = 36.33681 + 0.0046758 + X * 2.067302
X = 40.9574 amu
The isotopic mass of 41K is 40.9574 amu
Ten name if this compound is Potassium Oxide
Answer:
-125 kJ
Explanation:
You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.
H₂C=CH₂ + H₂ ⟶ H₃C-CH₃
Bonds: 4C-H + 1C=C 1H-H 6C-H + 1C-C
D/kJ·mol⁻¹: 413 612 436 413 347
The formula relating ΔHrxn and bond dissociation energies (D) is
ΔHrxn = Σ(Dreactants) – Σ(Dproducts)
(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)
<em>For the reactant</em>s:
Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ
<em>For the products:</em>
Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ
<em>For the system</em>
:
ΔHrxn = 2700 - 2825 = -125 kJ
