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.
Because it is less stable than phosphorous
Answer:
B)
Explanation:
This should be the correct answer, lmk if it's not
1 valence electron in alkali metals.
Answer: A) More base is likely required to reach the endpoint for the diprotic acid than for the monoprotic acid under these conditions
Explanation:
The monoprotic acid (HA) has a valency of 1 and diprotic acid
has a valency of 2.
As the concentration and volume of the diprotic acid and the monoprotic acids are equal.
The neutralization reaction for monoprotic acid is:

The neutralization reaction for diprotic acid is:

Thus more number of moles of base are required for neutralization of diprotic acid and thus the volume required will be more as concentration and volume of the diprotic acid and the monoprotic acids are equal.