As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
Answer: Yes we agree with the student's claim.
Explanation:
When the molecules are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules.Hence, more products will form leading to increased rate of reaction.
On increasing the temperature will make more reactant molecules will have sufficient energies to cross the energy barrier and thus the number of effective collisions increases, thus leading to more products and increased rate of reaction.
When the solution is stirred , the molecule's kinetic energy and thus the rate of reaction increases.
Thus smaller size, stirring and increase of temperature will make the solution quickly.
Answer:
11.39
Explanation:
Given that:
![pK_{b}=4.82](https://tex.z-dn.net/?f=pK_%7Bb%7D%3D4.82)
![K_{b}=10^{-4.82}=1.5136\times 10^{-5}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D10%5E%7B-4.82%7D%3D1.5136%5Ctimes%2010%5E%7B-5%7D)
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
![moles = \frac{Mass\ taken}{Molar\ mass}](https://tex.z-dn.net/?f=moles%20%3D%20%5Cfrac%7BMass%5C%20taken%7D%7BMolar%5C%20mass%7D)
Thus,
![Moles= \frac{1.805\ g}{82.0343\ g/mol}](https://tex.z-dn.net/?f=Moles%3D%20%5Cfrac%7B1.805%5C%20g%7D%7B82.0343%5C%20g%2Fmol%7D)
![Moles= 0.022\ moles](https://tex.z-dn.net/?f=Moles%3D%200.022%5C%20moles)
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
![Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%5C%20of%5C%20solute%7D%7BVolume%5C%20of%5C%20the%5C%20solution%7D)
![Molarity=\frac{0.022}{0.055}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B0.022%7D%7B0.055%7D)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)
![1.5136\times 10^{-5}=\frac {x^2}{0.4-x}](https://tex.z-dn.net/?f=1.5136%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%20%7Bx%5E2%7D%7B0.4-x%7D)
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
Answer:
In water hydrogen bonding formed between partially negative oxygen and partially positive hydrogen atoms.
Explanation:
Hydrogen bond is a partially electrostatic attraction between hydrogen (H), which is bonded to a more electronegative atom such as nitrogen (N), oxygen (O) or fluorine (F), and another adjacent atom bearing a lone pair of electrons.
In water molecule hydrogen and oxygen have difference in electronegativity, due to which dipoles are form. One is partial positive on hydrogen atom and other is partial negative on oxygen atom. The electrostatic attraction occur between two oppositely charged poles which is known as hydrogen bonding.