Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
Answer:
3.8 M
Explanation:
Volume of acid used VA= 57.0 - 37.5 = 19.5 ml
Volume of base used VB= 67.8 - 45.0 = 22.8 ml
Equation of the reaction
2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)
Number of moles of acid NA= 2
Number of moles of base NB= 1
Concentration of acid CA= ???
Concentration of base CB= 1.63 M
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA= 1.63 × 22.8 × 2/ 19.5 × 1
CA= 3.8 M
HENCE THE MOLARITY OF THE ACID IS 3.8 M.
Molar volume is when 1 mol of any gas occupies 22.4 L at STP.
Methyl ether has a mass of 8.12 g,
Volume occupied - 3.96 L
If 22.4 L occupied by 1 mol of gas
Then 3.96 L occupied by 1/22.4 x 3.96 = 0.176 mol of gas
The mass of 0.176 mol = 8.12 g
Molar mass is mass of 1 mol
Therefore mass of 1 mol = 8.12/0.176 = 46.1
Molecular weight is 46.1 g/mol
Explanation:
Answer is 4Al + 3O2 =>2Al2O3
I hope it's helpful!