Answer:
Ammonia > Urea > Ammonium nitrate > Ammonium sulphate
Explanation:
Percentage by mass of nitrogen in NH3:
Molar mass of NH3= 17 g/mol
Hence % by mass = 14/17 × 100 = 82.35%
% by mass of NH4NO3
Molar mass of NH4NO3 = 80.043 g/mol
Hence; 28/80.043 × 100 = 34.98%
% by mass of (NH4)2SO4;
Molar mass of (NH4)2SO4= 132.14 g/mol
Hence; 28/132.14 × 100 = 21.19%
% by mass of CH4N2O
Molar mass of urea = 60.0553 g/mol
Hence 28/60.0553 × 100 = 46.62%
Oxygen carbon and hydrogen
i believe its true bc ik for sure air is a homogenous mixture
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ