Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
Answer:
7
Explanation:
(Note: for a strong acid and strong base titration the equivalence point is at a pH=7. This is because at this point you have equal moles of added base as acid in the original solution. Therefore at the equivalence point the solution has formed a neutral salt and the pH is 7).
Answer:
0,1 mol
Explanation:
We know that the formula of concentration is C= moles of solute/ volume
0,4 M= moles of solute/ 250 mL
Convert mL to L 250 mL =0,25 L
0,4 M x 0,25 L= moles of solute
0,1 moles= moles of solute
Answer:
16.53 pounds
Explanation:
this type of problem needs convertion method we need to convert from grams to pound.
Explanation:
firstly find for the molar mass of kcl and molar mass of k
and then
molar mass of k = x
molar mass of kcl= 40
cross mutiply and then simplify you will get your answer
