<u>Given:</u>
The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J
The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J
<u>To determine:</u>
The final energy state Efinal of the electron
<u>Explanation:</u>
Since energy is being released, this suggests that Efinal < Einitial
i.e. ΔE = Einitial - Efinal
Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J
Ans: A)
The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J
Answer:
67.5%
Explanation:
Step 1: Write the balanced equation for the electrolysis of water
2 H₂O ⇒ 2 H₂ + O₂
Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O
According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.
17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂
Step 3: Calculate the percent yield of O₂
Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.
%yield = (exp yield / theoret yield) × 100%
%yield = (10.2 g / 15.1 g) × 100% = 67.5%
The molarity of KOH is 0.1055 M
<u><em> calculation</em></u>
Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH
H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O
step 2: find the moles of H₂C₂O₄.2H₂O
moles = mass÷ molar mass
from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol
= 0.2000 g ÷ 126 g/mol =0.00159 moles
step 3: use the mole ratio to calculate the moles of KOH
H₂C₂O₄.2H₂O : KOH is 1:2
therefore the moles of KOH =0.00159 x 2 = 0.00318 moles
step 4: find molarity of KOH
molarity = moles/volume in liters
volume in liters = 30.12/1000=0.03012 L
molarity is therefore = 0.00318/0.03012 =0.1055 M
Yes it is for example look at Iodine and Tellurium.
Hope this helps :).