density of acetone in gm/cm^3 is 0.791
=28.80cm^3*0.791 g/cm63
=22.78 g
Answer: a) pH of a 0.1 M vinegar solution is 2.9
b) It is an acid as pH is less than 7
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.
As vinegar is a weak acid, its dissociation is represented as;
cM 0 0
So dissociation constant will be:
Give c= 0.1 M
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.1\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.1%5Ctimes%20%5Calpha)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[1.3\times 10^{-3}]=2.9](https://tex.z-dn.net/?f=pH%3D-log%5B1.3%5Ctimes%2010%5E%7B-3%7D%5D%3D2.9)
Thus pH of a 0.1 M vinegar solution is 2.9
As pH is less than 7, it is an acid.
Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
