Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
A or D I think...sorry if it’s wrong
Answer:
P₂ = 20.5 torr
Explanation:
Given data:
Volume of tank = 15.0 L
Pressure of tank = 8.20×10⁴ torr
Final volume of tank = 6.00×10⁴ L
Final pressure = ?
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
8.20×10⁴ torr × 15.0 L = P₂ × 6.00×10⁴ L
P₂ = 8.20×10⁴ torr × 15.0 L / 6.00×10⁴ L
P₂ = 123×10⁴ torr.L/ 6.00×10⁴ L
P₂ = 20.5 torr