Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:

Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M

Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M


According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
It produces nitrogen gas and water
4NH3+3O2+heat 2N2+6H 2O
Depression in freezing point (Δ

) =

×m×i,
where,

= cryoscopic constant =

,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For

)
Thus, (Δ

) = 1.86 X 0.0085 X 2 =

Now, (Δ

) =

- T
Here, T = freezing point of solution

= freezing point of solvent =

Thus, T =

- (Δ

) = -
Answer:
0.05
moles
Explanation:
In a mole of any substance, there exist
6.02⋅1023
units of that substance.
So here, we got:
3.01⋅1022Mg atoms⋅1mol6.02⋅1023M gatoms=0.05mol