Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
That’s nice. Maybe bring your own next time
<span>The answer to your question is the 3rd option </span>
<span>1. The number of valence electrons increases as you go from left to right across a period. This is because the number of electrons increases, so another electron will be added to the outer shell.
2. Group 6A elements will have 6 valence elecrons, while Group 2A elements only have 2, therefore Group 6A elements have more valence electrons that Group 2A elements.
3. Fluorine has a smaller atomic size than the other halogens (Cl, Br, I), so its valence electrons are nearer to its nucleus. This means that the attractive forces are stronger, so when another electron (from another atom) draws near the F atom, it is more likely that the electron will be pulled toward the nucleus and react with the F atom</span>.
