Answer: 16.32 g of
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of
require = 1 mole of
Thus 0.34 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of
left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of
as excess reagent are left.
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
.
Note that water isn't part of this constant.
The value of
at 25 °C is
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
for pure water. - As a result,
at 25 °C.
Back to this question.
is given. 25 °C implies that
. As a result,
.
Answer:
56972.17K
Explanation:
P = 4.06kPa = 4.06×10³Pa
V = 14L
n = 0.12 moles
R = 8.314J/Mol.K
T = ?
We need ideal gas equation to solve this question
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
T = PV / nR
T = (4.06×10³ × 14) / (0.12 × 8.314)
T = 56840 / 0.99768
T = 56972.17K
Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa
The length of the seasons would increase