Answer:
Look at the pictures. On the 1 are compounds A and B. Compound c from b is on the 2nd image. Compound D is on 3rd image. Compound E is the same for compound C.
Explanation:
So for compound A sodium acetylide substitutes nucleophilicaly one Br on 1,12-dibromododecane. Then to obtain compound B sodium amide eliminates another Br. So for acetylene and alkene groups ozonolysis works the same way and we obtain diacid. Lyndlar catalyst works only on alkynes and make cis-alkenes from them. but we have a terminal alkyne for wich no isomers may occur. Pt reduction provides alkanes from both alenes and akynes. And sodium ammonia reduction works only on alkynes to provide trans-alkenes but, as I've said, isomers are not our case. So compounds E and C are the same and undergo same reaction with ozone.
In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
Answer:
Magnetic fields and Electric fields
Explanation:
Answer:
![H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}](https://tex.z-dn.net/?f=H%5E%2B%28aq%29%2B%28ClO%29%5E-%28aq%29%5Crightarrow%20HClO%7B%28aq%29%7D)
Explanation:
Hello there!
In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:
![2HClO_4{(aq)}+Ca(ClO)_2{(aq)}\rightarrow 2HClO{(aq)}+Ca(ClO_4)_2{(aq)}](https://tex.z-dn.net/?f=2HClO_4%7B%28aq%29%7D%2BCa%28ClO%29_2%7B%28aq%29%7D%5Crightarrow%202HClO%7B%28aq%29%7D%2BCa%28ClO_4%29_2%7B%28aq%29%7D)
Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:
![2H^++2(ClO_4)^-+Ca^{2+}+2(ClO)^-\rightarrow 2HClO{(aq)}+Ca^{2+}+2(ClO_4)^-](https://tex.z-dn.net/?f=2H%5E%2B%2B2%28ClO_4%29%5E-%2BCa%5E%7B2%2B%7D%2B2%28ClO%29%5E-%5Crightarrow%202HClO%7B%28aq%29%7D%2BCa%5E%7B2%2B%7D%2B2%28ClO_4%29%5E-)
Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:
![2H^+(aq)+2(ClO)^-(aq)\rightarrow 2HClO{(aq)}\\\\H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}](https://tex.z-dn.net/?f=2H%5E%2B%28aq%29%2B2%28ClO%29%5E-%28aq%29%5Crightarrow%202HClO%7B%28aq%29%7D%5C%5C%5C%5CH%5E%2B%28aq%29%2B%28ClO%29%5E-%28aq%29%5Crightarrow%20HClO%7B%28aq%29%7D)
Best regards!