NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27%
3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
0.8211 grams Na + 1.266 grams Cl = 2.087 grams
Answer:
c.) increase the concentration of one of the solutions
Explanation:
Explanation:
White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.
The chemical reaction is given as:
![AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)](https://tex.z-dn.net/?f=AgCl%28s%29%2B2NH_3%28aq%29%5Crightarrow%20Ag%5B%28NH_3%29_2%5D%5E%2BCl%5E-%28aq%29)
When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.
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.