<span>Water is considered as a polar molecule because its electrons has an uneven distribution. It has a partial negative charge and a partial positive charge on both ends of the molecule. </span>I hope my answer has come to your help. God bless and have a nice day ahead!
Answer:

Explanation:
When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.
So:
24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O
The next step is to divide each mass by their molar mass to convert your grams to moles.
24.42/40.08 = 0.6092 mol
17.07/14.01 = 1.218 mol
58.85/15.99 = 3.680 mol
Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.
0.6092 mol/0.6092 mol = 1
1.218 mol/0.6092 mol = 2
3.680 mol/0.6092 mol = 6
So the empirical formula is 
Answer:
Explanation:
The given reaction equation is:
2A + 4B → C + 3D
We know the mass of compound A in the reaction above. We are to find the mass of compound D.
We simply work from the known mass to calculate the mass of the unkown compound D
Using the mole concept, we can find the unknown mass.
Procedures
- We first find the molar mass of the compound A from the atomic units of the constituent elements.
- We then use the molar mass of A to calculate its number of moles using the expression below:
Number of moles of A = 
- Using the known number of moles of A, we can work out the number of moles of D.
From the balanced equation of the reaction, it is shown that:
2 moles of compound A was used up to produced 3 moles of D
Then
x number of moles of A would give the number of moles of D
- Now that we know the number of moles of D, we can find its mass using the expression below:
Mass of D = number of moles of D x molar mass of D
Answer: pH of resulting solution will be 13
Explanation:
pH is the measure of acidity or alkalinity of a solution.
Moles of
ion = 
Moles of
ion = 

For neutralization:
1 mole of
ion will react with 1 mole of
ion
0.01 mol of
ion will react with =
of
ion
Thus (0.012-0.01)= 0.002 moles of
are left in 20 ml or 0.02 L of solution.
![[OH^-]=\frac{0.002}{0.02L}=0.1M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B0.002%7D%7B0.02L%7D%3D0.1M)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![pOH=-log[0.1]=1](https://tex.z-dn.net/?f=pOH%3D-log%5B0.1%5D%3D1)


Thus the pH of resulting solution will be 13