Answer:
The charge on iodine is 2-
Answer: Radhika loves to eat sweets and chocolates. She asks her mother for the same before and after every meal. This is an example of bad eating habit.
Explanation:
Both sweets and chocolates contain high amount of fat and sugar. When Radhika is eating the same before and after every meal then it means she is consuming it in excess.
When we eat too much of sweets and chocolates then its remains get deposited in between our tooth and when they remain over there for hours then more amount of bacteria is formed.
Due to this bacteria our tooth starts to decay.
Hence, it is advised even by the doctors to eat sweets and chocolates occasionally and not regularly.
Thus, we can conclude that Radhika loves to eat sweets and chocolates. She asks her mother for the same before and after every meal. This is an example of bad eating habit.
When the compound PbI₂ dissolves, it dissociates as follows;
PbI₂ --> Pb²⁺ + 2I⁻
Molar solubility is the number of moles dissolved in 1 L of solution
A saturated solution is when the maximum amount of solute is dissolved in the solution.
Molar solubility of Iodide when solution is saturated is 2.7 x 10⁻³ mol/L, then solubility of Pb²⁺ is (2.7 x 10⁻³ mol/L) / 2 = 1.35 x 10⁻³ mol/L
ksp is the solubility product constant that can be calculated as follows;
ksp = [Pb²⁺][I⁻]
ksp = (1.35 x 10⁻³ mol/L) x (2.7 x 10⁻³ mol/L)²
= 1.35 x 10⁻³ x 7.29 x 10⁻⁶
= 9.8 x 10⁻⁹
Answer:
![[A_t]=0.037\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.037%5C%20M)
Explanation:
(a)
Given that:
Rate constant, k = 0.0347 min⁻¹
The expression for the half-life is:-
Where, k is rate constant
So,
(b)
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
t = 60 min
Initial concentration = 0.300 M
Final concentration =?
Applying in the above equation, we get that:-
![[A_t]=0.300\times e^{-0.0347\times 60}\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.300%5Ctimes%20e%5E%7B-0.0347%5Ctimes%2060%7D%5C%20M)
![[A_t]=0.3\times \frac{1}{e^{2.082}}\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.3%5Ctimes%20%5Cfrac%7B1%7D%7Be%5E%7B2.082%7D%7D%5C%20M)
4Li + O₂ = 2Li₂O
Li : Li₂O = 4 : 2 = 2 : 1
2 : 1
x : 4
x=4*2/1=8 mol
D) 8
