Guy-Lussac's Law states that the volume and the temperature are directly proportional given that the pressure remains constant.
For this problem, we will assume constant pressure. Based on the law:
(Volume/Temperatur)1 = (Volume/Temperature)2
(3.75/100) = (6.52/T)
T = 166.667 kelvin
Answer:
B: +3
Explanation:
If Gallium loses 3 electrons, it will become an ion.
The ion will be positively charged because in this new ion, the number of electrons is lesser than the number of protons. The charge difference will impart a positive net charge on the ion.
- In a neutral atom, the number of electrons and protons are the same.
- For positively charged ions, the number of protons is greater than the electrons
If Gallium the loss of 3 electrons offsets the charge balance in the chemical specie. Thus, the ion will have a net +3 charge.
A) - heat makes particles move faster, which usually makes the substance as a whole expand, but if the gas cannot expand the pressure will increase instead
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)
