Answer:
Iron; [Ar] 3d6 4s2
Cobalt; [Ar] 3d6 4s2 4p1
Explanation:
We know that in our universe, there are five d orbitals known. However, in this strange universe L, there are only three d orbitals known.
The sixth and seventh first row transition elements are iron and cobalt. In universe L, the electronic configuration of iron and cobalt will be written as;
Iron; [Ar] 3d6 4s2
Cobalt; [Ar] 3d6 4s2 4p1
The answer is Ba2+ + SO4 2- => BaSO4
Answer:
First, a mandatory stoichiometric balance must be made, and second, the degree of purities and impurities that the reagents have must be analyzed to see how many products they would have.
It is essential that this is taken into account since not all chemical compounds (reactants) are 100% pure and therefore their entirety does not become a product.
Explanation:
As for the stoichiometric balance, it is fundamental since if the equation is not balanced all the data will be wrong.
Many times the quantity in grams and another measurement of the products are obtained, taking into account a standard value of how many reactants will give x quantity of products, a simple rule of three could be made and thus be able to determine in a simpler way how many reactants were catalyzed in reaction.
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.
Under standard conditions :
E(cell) = E(cathode) - E(anode)
Note : cathode has the larger numeric value and anode has the smaller. Therefore
E(cell) = +1.36V - ( -3.04V)
= 1.36 + 3.04
= +4.40V
