Answer:
5. Selenium, because it does not have a stable, half-filled p subshell and adding an electron does not decrease its stability.
Explanation:
Electron affinity is the amount of energy released when an isolated gaseous atom accepts electron to form the corresponding anion.
Selenium:-
The electronic configuration of the element is:-
![[Ar]3d^{10}4s^24p^4](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E24p%5E4)
Arsenic:-
The electronic configuration of the element is:-
![[Ar]3d^{10}4s^24p^3](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E24p%5E3)
The 4p orbital in case of arsenic is half filled which makes the element having more stability as compared to selenium.
Thus, selenium has higher electron affinity because adding electron does not decrease the stability as in case of arsenic.
Answer:
C24H50
Explanation:
The empirical fomula's molar mass is 169.25 g/mol.
We know the molecular formula's molar mass is 338 g/mol.
338/169.25= 1.99 or approximately 2
Answer:
0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.
Explanation:
When hydrogen and nitrogen react they form ammonia.
Chemical equation:
N₂ + 3H₂ → 2NH₃
Given mass of nitrogen = 2.75 g
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2.75 g / 28 g/mol
Number of moles = 0.098 mol
Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:
N₂ : H₂
1 : 3
0.098 : 3×0.098 = 0.3 mol
Mass of hydrogen:
Mass = number of moles × molar mass
Mass = 0.3 mol × 2 g/mol
Mass = 0.6 g
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)