Cambium - <span>a cellular plant tissue from which phloem, xylem, or cork grows by division, resulting (in woody plants) in secondary thickening.</span>
Answer:
Complete ionic:
.
Net ionic:
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water:
,
, and
. These three salts will exist as ions:
- Each
formula unit will exist as one
ion and one
ion. - Each
formula unit will exist as one
ion and two
ions (note the subscript in the formula
.) - Each
formula unit will exist as one
and two
ions.
On the other hand,
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite
,
, and
(three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each
formula unit will exist as only one
ion and one
ion. However, because the coefficient of
in the original equation is two,
alone should correspond to two
ions and two
ions.
Do not rewrite the salt
because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of
and two units of
. Doing so will give:
.
Simplify the coefficients:
.
X

H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0
Y</span>

is also neutral so 2X+Y= 0
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge
Answer: The valency of X is -3.
The valency of Y is 6
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
An exchange reaction consists of both synthesis and decomposition reactions.
Here’s a complex example: AB + CD → AC + BD.
Another example might be: AB + CD → AD + BC.