Answer:
Examples of Chemical Changes
Burning wood.
Souring milk.
Mixing acid and base.
Digesting food.
Cooking an egg.
Heating sugar to form caramel.
Baking a cake.
Rusting of iron.
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:
The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:
Expression for is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Answer:
C) It has a constant average kinetic energy
Explanation:
The average kinetic energy of the particles in a gas is directly proportional to the temperature of the gas, according to the equation.
k is the Boltzmann's constant
T is the absolute temperature of the gas
Therefore, temperature of a gas is a measure of the average kinetic energy of the particles.
In this problem, we are told that the gas is at constant temperature (and volume): therefore, according to the previous equation, this means that the average kinetic energy is also constant.
Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
The balanced reaction equation for the reaction between CH₃OH and O₂ is
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles 12 24
Reacted moles 12 18
Final moles - 6 12 24
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
= 12 mol x 3 / 2
= 18 mol
All of CH₃OH moles react with O₂.
Hence, the limiting agent is CH₃OH.
Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
