Answer:
d) He, Ar, Kr
Explanation:
To have similar chemical properties, elements must belong to the same period on the periodic table.
As a generally periodic trend, elements within the same group, have the same chemical properties.
So, the correct answer choice will draft elements from the same group.
- He, Ar and Kr belongs to the group of elements called the noble gases.
- They are an unreactive group of elements.
- They are found in the last group on the periodic table.
- These elements do not readily combine with other elements.
The best option is melting point
The reason for this is that <span>these two ketones are so small that they have only one possible ketone. So the number is usually omitted. Normally the ketone group needs a number but these two are exceptions</span>
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Answer:
H2 + O2
Explanation:
H2 + O2 --> H2O
The items on the left side is the reactants. The arrow means yields, and the item the arrow is pointing at is the product.
