Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
Answer:
Software Developer. ...
Database Administrator. ...
Computer Hardware Engineer. ...
Computer Systems Analyst. ...
Computer Network Architect. ...
Explanation:
Answer:
Cellular respiration.
Explanation:
Through the process of cellular respiration, the energy in food is converted into energy that can be used by the body's cells. During cellular respiration, glucose and oxygen are converted into carbon dioxide and water, and the energy is transferred to ATP.
Answer: 35 g/cm
Explanation:
Density equals mass over volume. 525 divided by 15 is 35
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample =
= 165°C
Depression in freezing point = ![\Delta T_f=?](https://tex.z-dn.net/?f=%5CDelta%20T_f%3D%3F)
![\Delta T_f=T- T_f=179^oC-165^oC=14^oC](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DT-%20T_f%3D179%5EoC-165%5EoC%3D14%5EoC)
Depression in freezing point is also given by formula:
![\Delta T_f=i\times K_f\times m](https://tex.z-dn.net/?f=%5CDelta%20T_f%3Di%5Ctimes%20K_f%5Ctimes%20m)
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have:
= 40°C kg/mol
i = 1 ( organic compounds)
![\Delta T_f=14^oC](https://tex.z-dn.net/?f=%5CDelta%20T_f%3D14%5EoC)
![14^oC=1\times 40^oC kg/mol\times m](https://tex.z-dn.net/?f=14%5EoC%3D1%5Ctimes%2040%5EoC%20kg%2Fmol%5Ctimes%20m)
![m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B14%5EoC%7D%7B1%5Ctimes%2040%5EoC%20kg%2Fmol%7D%3D0.35%20mol%2Fkg)
The molality of isoborneol in camphor is 0.53 mol/kg.