Solution here,
Volume(V)=67.4 L
Pressure(P)=1 atm
Temperature(T)=(0+273)K=273K
Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1
No. of moles(n)=?
Now,
PV=nRT
or, 1×67.4=n×0.0821×273
or, 67.4=22.4n
or, n=67.4/22.4
or, n=3
therefore, required no. of mole is 3.
I think it is D I'm not positive
Hi there!
Zinc: Is qualitative
Chlorine: is quantitative
Gallium: is neither
Nitrogen: is quantitative
Aluminum: is quantitative
If you need an explanation, please let me know !
Hope this helps and have a good day :) !
~Angel
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y
3y = 1.2
y = 0,4M
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4
C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄
mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol
164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄