Z=1 is the formula i would have to see the following lol
Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄
(1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
![Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B-2%7D%5D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5D%20%5BH_%7B2%7DO%5D%7D%20%3D%206.2x10%5E%7B-8%7D)
The pKa is

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.
Answer:
C) SN2 and E2
Explanation:
For this question, we have analyzed the <u>substrate</u> and the <u>base/nucleophile</u>. The substrate, in this case, is 1-iodohexane and the base/nucleophile is potassium tert-butoxide.
<u>Substrate</u>
<u />
In the 1-iodohexane the iodide "I" is bonded to a primary carbon (carbon 1). Therefore we will have a <u>primary substrate</u>. If we have a primary substrate an Sn1 can not take place. We can not have a <u>primary carbocation</u> due to this instability. So, we can disccard options A) and B).
<u>Base/nucleophile</u>
<u />
In the potassium tert-butoxide we have an ionic compound. A positive charge is placed in the potassium atom a negative charge is placed in the oxygen of the ter-butoxide ion. So, we will have a <u>strong base</u> (a molecule with the ability to remove electrons) and a <u>strong nucleophile</u> (a molecule with ability to bond with an electrophile). With all this in mind, w<u>e can not have an E1 reaction</u>.
With both analyses, the answer is C).
See figure 1
I hope it helps!
Answer:
Metallic bonding
Explanation:
Metals have low ionization energies. Therefore, their valence electrons are easily delocalized (attracted to the neighbouring metal atoms). These delocalized electrons are then not associated with a specific metal atom. Since the electrons are “free”, the metal atoms have become cations, and the electrons are free to move throughout the whole crystalline structure.
We say that a metal consists of an array of cations immersed in a sea of electrons
.
The electrons act as a “glue” holding the cations together.
Metallic bonds are the attractive forces between the metal cations and the sea of electrons.
In an NaK alloy, for example, the Na and K atoms contribute their valence electrons to the "sea". The atoms aren’t bonded to each other, but they are held in place by the metallic bonding.
A is true of UV rays.
B is true not of UV rays but rather of visible light.
C is true not of UV rays but rather of microwaves. (unless you actually toast your toast in a toaster like a normal person)
D is true not of UV rays but rather of radio waves.