Answer:
c. The reaction will proceed rapidly from left to right.
Explanation:
The variation of the free Gibbs energy doesn't tell anything about the speed of reaction.
On the other hand, when ΔGo is negative: the reaction is spontaneous, thermodynamically favourable, and the products are more stable than the reactants
Answer:
1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵
Explanation:
1) Ca:
Its atomic number is 20. So it has 20 protons and 20 electrons.
Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.
So, you have two more electrons (20 - 18 = 2) to distribute.
Those two electrons go the the orbital 4s.
Finally, the electron configuration is [Ar] 4s².
2) Pm
The atomic number of Pm is 61, so it has 61 protons and 61 electrons.
Pm is in the row (period) 6. So, the noble gas before Pm is Xe.
The atomic number of Xe is 54.
Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.
The resultant distribution for Pm is: [Xe]6s² 4f⁵.
Answer:
i am pretty sure the answer is a
Explanation: because the airplane's flight time has to be the independent variable for it to affect the dependent variable that is the speed of how fast the airplane is going.
Answer:
![[Ag^{+}]=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
Explanation:
Given:
[AgNO3] = 0.20 M
Ba(NO3)2 = 0.20 M
[K2CrO4] = 0.10 M
Ksp of Ag2CrO4 = 1.1 x 10^-12
Ksp of BaCrO4 = 1.1 x 10^-10
![Ksp=[Ba^{2+}][CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-10%7D%3D%280.20%29%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}](https://tex.z-dn.net/?f=%5BCrO_%7B4%7D%5E%7B2-%7D%5D%3D%5Cfrac%7B1.2%5Ctimes%2010%5E%7B-10%7D%7D%7B%280.20%29%7D%3D%206.0%5Ctimes%2010%5E%7B-10%7D)
Now,
![Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})](https://tex.z-dn.net/?f=1.1%5Ctimes%2010%5E%7B-12%7D%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%286.0%5Ctimes%2010%5E%7B-10%7D%29)
![[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%3D%5Cfrac%7B1.1%5Ctimes%2010%5E%7B-12%7D%7D%7B%286.0%5Ctimes%2010%5E%7B-10%7D%29%7D%3D%201.8%5Ctimes%2010%5E%7B-3%7D)
![[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-3%7D%7D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M
