2Cu(NO3)2 ----------> 2CuO (s) +4NO2 (g) + O2(g)
9.378g=0.05moles
no of moles = weight / MW = 9.378/187.56 = 0.05moles
as per the above reaaction 2moles of Cu(NO3)2 can produce 4moles of N2
0.05moles Cu(NO3)2 can produce (0.05*4)/2 = 0.1moles of N2
and 2moles of Cu(NO3)2 can produce 1moles of O2
0.05moles Cu(NO3)2 can produce (0.05*1)/2 = 0.025moles of O2
Total moles of gas i.e., N2 and O2 =0.1+0.025 = 0.125moles
From PV = nRT
V = nRT/ P = 0.125*0.0821*273 = 2.80166Lit option is correct
Answer:
S = 21.92 %
F = 78.08 %
Explanation:
To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.
Molar mass of SF6 = 32 + 19(6)
= 32 + 114
= 146g/mol
mass of Sulphur (S) in SF6 = 32g
mass of Fluorine (F) in SF6 = 114g
Percent composition = mass of element/molar mass of compound × 100
- % composition of S = 32/146 × 100 = 21.92%.
- % composition of F = 114/146 × 100 = 78.08%.
hope it helps ..............
Magnet to remove the iron, filter paper (or some filter that could trap sand) to remove the sand, and heat up the salt and water, causing the water to evaporate to separate them.
The mass of the liquid is 4.8 kg.
6.4 L = 6400 mL
Mass = 6400 mL × (0.70 g/1 mL) = 4500 g = 4.8 kg
