You can boil or evaporate the water and the salt will be left behind as a solid. If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid. So, I would say maybe oil.
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
![pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247](https://tex.z-dn.net/?f=pKa%3D-log%5BH%5D%20%3D%20-%20log%20%5B%205.66%20%2A%2010%5E%7B-7%7D%5D%5C%5C%20%5C%5Cpka%20%3D%207%20-%20log%20%285.66%29%3D7-0.753%3D6.247%5C%5C%5C%5Cpka%20%3D%206.247)
The pH of the buffer can be known as
![pH = pK_{a} + log[\frac{[A-]}{[HA]}}]](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5B%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%7D%5D)
The concentration of ![[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20Moles%20of%20%5BA%5D%2FTotal%20volume%20%3D%200.608%2F2%20%3D%200.304%20M%5C%5C)
Similarly, the concentration of [HA] = 
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]

So, the pH of the buffer is 6.1236.
if your serious about this question then it is 5
1) is right
2) lead(ii) phosphate
3) iron(iii)sulfate
4) lead(ii)oxide
5) lead sulfate
6) copper iodide
7) lead oxide
<u>Answer:</u> The temperature when the volume and pressure has changed is 274 K
<u>Explanation:</u>
To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:

where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
![P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K](https://tex.z-dn.net/?f=P_1%3D760mmHg%5C%5CV_1%3D175L%5C%5CT_1%3D15%5EoC%3D%5B15%2B273%5DK%3D288K%5C%5CP_2%3D640mmHg%5C%5CV_2%3D198L%5C%5CT_2%3D%3FK)
Putting values in above equation, we get:

Hence, the temperature when the volume and pressure has changed is 274 K