CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Explanation:
As a neutral lithium atom contains 3 protons and its elemental charge is given as 
. Hence, we will calculate its number of moles as follows.
Moles = 
= 
= 100 mol
According to mole concept, there are 
atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.
No. of atoms = 
= 
atoms
Since, it is given that charge on 1 atom is as follows.
= 
Therefore, charge present on atoms will be calculated as follows.
Thus, we can conclude that a positive charge of is in 0.7 kg of lithium.
Answer:
V = 34430 mL
Explanation:
Given data:
Volume in mL = ?
Number of moles of gas = 2.00 mol
Temperature = 36°C (36+273= 309K)
Pressure of gas = 1120 torr
Solution:
Formula:
PV = nRT
V = nRT/P
V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr
V = 38563.2 torr • L / 1120 torr
V = 34.43 L
L to mL
34.43 L ×1000 mL / 1 L
34430 mL
Find the hydroxide ion concentration of a solution with a pOH of 5.90. To solve this, use a scientific calculator and enter 5.90 and use the +/- button to make it negative and then press the 10x key.
