pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Answer:
The amount of Chlorodecane in the unknown is 0.105nmols
Explanation:
a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.
The area of the first peak corresponding to Chlorohexane is 32434 units.
The area of the second peak corresponding to chlorodecane is 2022 units.
Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:
1.69 nmols of Chlorohexane gives 32434 units
How much of chlorodecane gives 2022 units
By cross multiplication;
Moles of Chlorodecane = 2022*1.69/32434
=0.105nmols
Answer:
A. Coefficients
Explanation:
that's the number in front of the molecules
Answer:
Nitrogen
Oxygen
Argon
Carbon Dioxide
Methane
Ozone
Explanation:
N₂ accounts for 78% of the atmosphere.
O₂ accounts for 21% of the atmosphere.
Ar accounts for 0.9% of the atmosphere.
CO₂, CH₄, and O₃ only take up 0.1% of the atmosphere.
