What do you need help with
Answer:
a )Li
b)O
c)F
Explanation:
a) Li-1s^2 2s^1
F-1s^2 2s^2 2p^5
it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy
b)oxygen ion is larger than Na because o have fewer proton
c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.
Answer:
6.46 × 10⁻¹¹ M
Explanation:
Step 1: Given data
pH of the solution: 3.81
Step 2: Calculate the pOH of the solution
We will use the following expression.
pH + pOH = 14.00
pOH = 14.00 - pH = 14.00 - 3.81 = 10.19
Step 3: Calculate the concentration of OH⁻ ions
We will use the definition of pOH.
pOH = -log [OH⁻]
[OH⁻] = antilog -pOH = antilog -10.19 = 6.46 × 10⁻¹¹ M
Answer:
The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside.
Explanation: