What is the concentration of 20 mL of LiOH if it is neutralized by 60 mL of 4 M HCl?

The concentration is 12M

Step-by-step explanation:

In this question, we are asked to calculate the concentration of the alkali given the information in the question.

To answer this question correctly, the first thing to do is to write the complete chemical equation of the reaction of the acid and the alkali. We identify that LiOH is the alkali and HCl is the acid

LiOH + HCl ———-> LiCl + H2O

we can deduce that 1 mole of the acid used up a mole of the base.

Now, we use the titration formula;

CaVa/Na. = CbVb/Nb

since Na= Nb = 1 i.e number of moles , we can just write the formula to be

CaVa. = CbVb

Where Ca is the concentration of the acid which is 4M

Va is the volume of acid used which is 60mL

Cb is the concentration of the base which is ?

Vb is volume of base which is 20mL

Plugging these values, we have;

4 * 60 = Cb * 20

Cb = 240/20

Cb = 12M

The concentration of the base is thus 12M

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement