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:
.
Answer:
During MITOSIS, the parent, diploid (2n), cell is divided to create two identical, diploid (2n), daughter cells. ... After cytokinesis, the ploidy of the daughter cells remains the same because each daughter cell contains 4 chromatids, as the parent cell did.
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
The type of bonds present in the compound. and the type of structure it has and the elements that are presents and the number of moles of each element in one mole of the compound.
