Answer:
<h3>601.93 g/mol</h3>
<h3>explanation:</h3>
Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.
Answer:
0.1
Explanation:
We must first put down the equation of the reaction in order to guide our solution of the question.
2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)
Now from the question, the following were given;
Concentration of acid CA= ??????
Concentration of base CB= 0.299M
Volume of acid VA= 17.8ml
Volume of base VB= 24.7ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
SUBSTITUTING VALUES;
CA= 0.299 × 24.7 ×2 / 17.8×1
CA= 0.8298 M
But;
pH= -log[H^+]
[H^+] = 0.8298 M
pH= -log[0.8298 M]
pH= 0.1
D. The empirical formula and the molar mass
E = HF, where H is Planck's constant, 6.63 x 10 - 34 j.s
I’m pretty sure it’s B
Hope that helps
