You can make a solution saturated by Adding more solute. Hope this helps, good luck.
The solubility of gas in water is inversely proportional to the temperature. That means cool waters can hold more gases than hot waters. So when the oceans continue to warm all the green--houses gases present in oceanic waters will be released into to the atmosphere. This would further lead to the heating up of the planet. The global climate would keep changing and the temperature of the planet would increase further. Therefore, when the oceans continue to warm the amount of green-house gases cannot be sequestered by the oceans (as the temperatures are high) and so this would further enhance the greenhouse effect.
The purpose of using distilled water in the cuvette of the spectrophotometer is to calibrate the instrument. It is generally considered that distilled water shows zero absorbance and 100 % transmittance. So, to zero out the absorbance of compounds other than the analyte being determined, distilled water is used as a blank.
If the unknown being determined is prepared using ethanol as the base solution, the blank used must be ethanol. This is because absorbance if any from the solvent, ethanol must be zeroed out as when the measurement of the actual unknown is being made, the absorbance of the solvent does not interfere.
The charge of a Rb ion would be +1
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54