Since the leaving ability of the halide ions increasees as the basicity of the halide decreases.
If the basicity of the halide decreases as the its conjugate acid is strong.
Since the pKa value of conjuage acid of haldie is 3, it is a weak acid. So, it halide is not a good leaving group.
Therefore the answer is No, because a good leaving group is the conjugate base of a strong acid. The halide not acidic enough to be a good leaving group.
Answer:
hypertonic solution
Explanation:
Hypertonic solution -
It is the solution, with more amount of solute than the solvent , is known as hypertonic solution.
Now, is some substance is immersed in such solution , the substance gets shrinked , because , the solvent from the substance moves out of it and moves to the hypertonic solution.
Hence, the pickles gets shrinked up , as they put in a hypertonic solution.
Answer:
Incomplete question: 9.9 mL of 0.01 M EDTA, and 10 mL of buffer with a pH of 4.
The concentration of [HY³⁻] is 1.8x10⁻²⁸mol/L
Explanation:
Given information:
10 mL of 0.01 M VOSO₄ = 0.01 L
9.9 mL of 0.01 M EDTA = 0.0099 L
10 mL of buffer = 0.01 L
pH of 4
Question: Calculate HY³⁻, [HY³⁻] = ?
Moles of EDTA:
![n_{EDTA} =0.01\frac{moles}{L} *0.0099L=9.9x10^{-5} moles](https://tex.z-dn.net/?f=n_%7BEDTA%7D%20%3D0.01%5Cfrac%7Bmoles%7D%7BL%7D%20%2A0.0099L%3D9.9x10%5E%7B-5%7D%20moles)
The concentration of Y⁴⁺:
At pH = 4 Y⁴⁻ = 3.8x10⁻⁹
mol/L
The concentration of [H₂Y²⁻]:
![[H_{2} Y^{2-} ]=Kf*[VO_{2} ^{2+} ][Y^{4+} ]=6.9x10^{-7} *(0.01*0.01)*3.762x10^{-13} =2.6x10^{-23}](https://tex.z-dn.net/?f=%5BH_%7B2%7D%20Y%5E%7B2-%7D%20%5D%3DKf%2A%5BVO_%7B2%7D%20%20%5E%7B2%2B%7D%20%5D%5BY%5E%7B4%2B%7D%20%5D%3D6.9x10%5E%7B-7%7D%20%2A%280.01%2A0.01%29%2A3.762x10%5E%7B-13%7D%20%3D2.6x10%5E%7B-23%7D)
The reaction:
H₂Y²⁻ → H⁺ + HY³⁻
The concentration of HY³⁻
![6.9x10^{-7} =\frac{[H+][HY^{3-}] }{[H_{2}Y^{2-}] } \\6.9x10^{-7} =\frac{0.1*[HY^{3-}]}{2.6x10^{-23} }](https://tex.z-dn.net/?f=6.9x10%5E%7B-7%7D%20%3D%5Cfrac%7B%5BH%2B%5D%5BHY%5E%7B3-%7D%5D%20%7D%7B%5BH_%7B2%7DY%5E%7B2-%7D%5D%20%20%7D%20%5C%5C6.9x10%5E%7B-7%7D%20%3D%5Cfrac%7B0.1%2A%5BHY%5E%7B3-%7D%5D%7D%7B2.6x10%5E%7B-23%7D%20%7D)
Solving for HY³⁻
[HY³⁻] = 1.8x10⁻²⁸mol/L
Answer:
B
Explanation:
The last 2 answers can immediately be cancelled out as they're not combustion of ethanol, combustion in this context is not a reversable reaction so cannot be referenced in this way.
The answer is determined to be B just by simply counting the numbers of each of the elements on the reactants (left) side and the products (right) side and ensuring they're the same
In A on the reactants side there is 2C 6H 3O and on the products side there is 1C 3O and 2H its not equal so it is not correct
In B on the reactants side there is 2C 6H 7O and on the products side there is 2C 6H 7O its equal so this answer is correct
Answer:
<span>When identical particles pack in a simple cubic lattice, there is ONE particle per unit cell.
Explanation:
As we know that simple cube has eight corners and each particle at every corner is shared by eight cubic crystals. So it means that only </span><span>eighth </span>part of a particle is present at one corner of each cube. Therefore, each corner contain 1/8 particle, then all corners will contain 1/8 × 8 = 1 particle.