The energy transformation occurs in a gasoline-powered car is c<span>hemical energy is converted into mechanical energy. The answer is letter A. The rest of the choices do not answer the question above.</span>
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
They are in order according to their atomic number, so the position should tell you the atomic number.
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
One is through sublimation, where the mixture is heated and iodine gets converted into gaseous form, leaving behind the iron fillings.
The other is to get a magnet near the mixture and all the iron fillings get attracted to it while iodine will be left over
