Answer:
The interaction between sodium ion and the partial negative charge on the oxygen is stronger.
Explanation:
The predominant interaction that exists between sodium ion and the partial negative charge on the oxygen is ion-dipole interaction.
The predominant interaction that exists between two ethanol molecules is hydrogen bonding interaction.
The order of strength of the intermolecular interactions in decreasing order:
Ionic bond > ion-dipole interaction > hydrogen bonding > dipole-dipole interaction > ion-induced dipole interaction > induced dipole- dipole interaction > london force
So, ion-dipole interaction is stronger than hydrogen bonding.
Hence, the interaction between sodium ion and the partial negative charge on the oxygen is stronger.
The energy generated by the movement of electrons is used to pump electrons across the inner mitochondrial membrane to an area of higher concentration. 17. Where do these protons (H+) come from? The originally came from a glucose molecule and were carried to the electron transport chain by NADH and FADH2.
The answer would be glucose molecule.
Answer:
The correct answer is option a.
Explanation:
When aluminum hydroxide reacts with of nitrous acid it gives of aluminum nitrite and of water.
According to above reaction ,when 1 mole of aluminum hydroxide reacts with 3 moles of nitrous acid it gives 1 mole of aluminum nitrite and 3 moles of water.
Hence, the correct answer is option a.
The reaction is actually endothermic because delta H is positive, indicating that it absorbing heat.
Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)
First step. Need to know how much moles of the substances are present
![0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH](https://tex.z-dn.net/?f=0.1%20%5Cfrac%7Bmol%20NaOH%7D%7BL%7D%20%2A%200.030%20L%20%3D%200.003%20mol%20NaOH%3C%2Fp%3E%3Cp%3E%5Btex%5D0.1%20%5Cfrac%7Bmol%20CH_3COOH%7D%7BL%7D%20%2A%200.025%20L%3D%200.0025%20mol%20CH_3COOH%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ESecond%20step.%20Know%20wich%20substance%20is%20in%20excess.%3C%2Fp%3E%3Cp%3E0.0025%20mol%20CH_3COOH%20%2A%20%5Btex%5D1%20mol%20NaOH%2F%201%20mol%20CH_3COOH)
= 0.0025 mol NaOH
0.003 mol NaOH * 
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
