Answer:

Explanation:
The products of this reaction are given by:

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

The following steps should be taken:
- balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

- balance oxygen atoms by adding 7 water molecules on the right:

- balance the hydrogen atoms by adding 14 protons on the left:

- balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

Similarly, tin(II) cation becomes tin(IV) cation:

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:


Add them together:

Adding the ions spectators:

Answer:
At the end of cell cycle (mitosis replication), two daughter cells are produced. Each contains exactly the same number of DNA content.
Answer:
✓ scholastics
Explanation:
you d.ont need a expla.nation rig.ht un.less y.ou wan.na re.ad for an h.our
Effect of increasing surface area on the rate of a reaction. ... Increasing the surface area of a solid reactant exposes more of its particles to attack. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.