The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is 
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 
Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
The shape of the H2O molecule is a Bent Triatomic.
It isn't symmetrical.
The H2O molecule is polar.
Answer:
The density of acetic acid at 30°C = 1.0354_g/mL
Explanation:
specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)
Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)
= 0.9956 g/mL × 1.040
= 1.0354_g/mL
Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.
Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure
Answer:G
Explanation:
I would look it up on safari just to be sure tho, hope this helps
The molar mass of the unknown gas is 184.96 g/mol
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>How to determine the molar mass of the unknown gas </h3>
The following data were obtained from the question:
- Rate of unknown gas (R₁) = R
- Rate of CH₄ (R₂) = 3.4R
- Molar mass of CH₄ (M₂) = 16 g/mol
- Molar mass of unknown gas (M₁) =?
The molar mass of the unknown gas can be obtained as follow:
R₁/R₂ = √(M₂/M₁)
R / 3.4R = √(16 / M₁)
1 / 3.4 = √(16 / M₁)
Square both side
(1 / 3.4)² = 16 / M₁
Cross multiply
(1 / 3.4)² × M₁ = 16
Divide both side by (1 / 3.4)²
M₁ = 16 / (1 / 3.4)²
M₁ = 184.96 g/mol
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
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