Answer : The concentration of and at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M
As molecular mass increases, van der Waals interactions between molecules also increase, because numbers of electrons in the molecules.
Iodine (I) has greatest atomic number, so the number of electrons.
Hydrogen fluoride is exception, because hydrogen bonding between molecules.
Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.
Explanation:
First convert the amount of water into moles:
360 g H2O ×
Now let's calculate the number of moles of O2 gas produced.
20 mol H2O ×
The volume of gas at 10°C and 5 atm can be found using the ideal gas law:
Answer: is the limiting reagent
Explanation:
To calculate the moles :
The balanced chemical reaction is :
According to stoichiometry :
1 moles of require = 2 moles of
Thus 0.625 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present more than the required amount.