The first thing to do is to calculate the new concentration of the solution after dilution. We do as follows:
M1V1 = M2V2
(3.0 M).006 L = M2 (18 L)
M2 = 0.001 M HNO3
When in solution it dissociated into ions as follows:
HNO3 = H+ + NO3-
[H+] = 0.001 M HNO3 ( 1 mol H+ / 1 mol HNO3 ) = 0.001 M H+
pH = -log[H+] = -log 0.001 = 3
I think it’s A. Subtropical Monsoon
2.1653 g
Explanation:
The molar mass of Rubidium is;
85.468 g/mol
Therefore the moles of Rubidium that reacted with oxygen is;
1.98 / 85.468
= 0.0232 moles
If every two moles of Rubidium reacts with one mole of oxygen then the amount of oxygen consumed in the chemical reaction is;
0.5 * 0.0232
= 0.0116 moles
The molar mass of an oxygen atom is 16 g/mole. Then the amount of O in grams consumed is;
0.0116 * 16
=0.1853 g
The final weight of the Rubidium II Oxide is;
1.98 + 0.1853
= 2.1653 g
Answer:
peroid num= 3
group num=13 (boron Group)
Explanation:
