Answer: 50%
Explanation:
The number of electron pairs are 2 for hybridization to be
and the electronic geometry of the molecule will be linear.
1. percentage of s character in sp hybrid orbital =
2. percentage of s character in
hybrid orbital =
3. percentage of s character in
hybrid orbital =
Thus percentage of s-character in an sp hybrid is 50%.
Answer:
During MITOSIS, the parent, diploid (2n), cell is divided to create two identical, diploid (2n), daughter cells. ... After cytokinesis, the ploidy of the daughter cells remains the same because each daughter cell contains 4 chromatids, as the parent cell did.
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
Answer:
B, D, E, C, A
Explanation:
We have 5 blocks with their respective masses and volumes.
Block Mass Volume
A 65.14 kg 103.38 L
B 0.64 kg 100.64 L
C 4.08 kg 104.08 L
D 3.10 kg 103.10 L
E 3.53 kg 101.00 L
The density (ρ) is an intensive property resulting from dividing the mass (m) by the volume (V), that is, ρ = m / V
ρA = 65.14 kg / 103.38 L = 0.6301 kg/L
ρB = 0.64 kg / 100.64 L = 0.0064 kg/L
ρC = 4.08 kg / 104.08 L = 0.0392 kg/L
ρD = 3.10 kg / 103.10 L = 0.0301 kg/L
ρE = 3.53 kg / 101.00 L = 0.0350 kg/L
The order from least dense to most dense is B, D, E, C, A