M(KNO₃)=101.1 g/mol
M(CO(NH₂)₂)=60.1 g/mol
m(N)=M(N)m(KNO₃)/M(KNO₃)
m(N)=2M(N)m(CO(NH₂)₂)/M(CO(NH₂)₂)
2m(CO(NH₂)₂)/M(CO(NH₂)₂)=m(KNO₃)/M(KNO₃)
m(CO(NH₂)₂)=M(CO(NH₂)₂)m(KNO₃)/(2M(KNO₃))
m(CO(NH₂)₂)=60.1*101.1/(2*101.1)=30.05 g
Both bc your buying the product and a service of them installing it
Answer:
Explanation:
Entropy is measure of disorder so as we lower the temperature of gas , its entropy decreases .
Hence at - 200°C entropy of nitrogen will be less than that at - 190°C .
At freezing point ,
entropy of fusion = latent heat / freezing temperature
= .71 kJ / ( 273 - 210 )
= 710 / 63 J mol⁻¹ K⁻¹ .
= 11.27 J mol⁻¹ K⁻¹ .
entropy of fusion = 11.27 J mol⁻¹ K⁻¹ .
The force of the swing.
Hoped I helped, and please tell me if I am wrong.
