The solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L
<h3>Effect of Pressure on Solubility </h3>
As the <em>pressure </em>of a gas increases, the <em>solubility </em>increases, and as the <em>pressure </em>of a gas decreases, the <em>solubility </em>decreases.
Thus, Solubility varies directly with Pressure
If S represents Solubility and P represents Pressure,
Then we can write that
S ∝ P
Introducing proportionality constant, k
S = kP
S/P = k
∴ We can write that
Where is the initial solubility
is the initial pressure
is the final solubility
is the final pressure
From the given information
Putting the parameters into the formula, we get
Hence, the solubility of the gas if the pressure is changed to 100 kPa is 0.742 g/L
Learn more on Solubility here: brainly.com/question/4529762
Answer: I’m guessing ibhs and mr Kadda right?
Explanation:I’m stuck too
The number of moles is given by:
Mass of hydrogen, = (given)
Molar mass of hydrogen, =
Number of moles of hydrogen, =
Reaction between to form is:
From the reaction it is clear that 3 moles of hydrogen gives 2 moles of ammonia and 1 mole of nitrogen gives 2 moles of ammonia .
So, of gives:
Number of moles of ammonia, =
Since, in one mole of ammonia, there are molecules of ammonia, .
So, number of molecules in of ammonia, is:
Hence, the number of molecules of ammonia produced by of is .
Answer:
The answer to your question is: 1.205 x 10²³ atoms of Carbon
Explanation:
Data
Carbons atoms contained in 2.8 g C₂ H₄
MW C₂ H₄ = (12 x 2) + (4 x 1) = 24 + 4 = 28 g
28 g of C₂ H₄ ---------------- 24 g of carbon
2.8 g C₂ H₄ ----------------- x
x = (2.8)(24) / (28)
= 2.4 grams of C
1 mol ------------------- 12 g of Carbon
x ------------------- 2.4 g
x = (2.4x 1) / 12 = 0.2 mol
1 mol ----------------- 6.023 x 10 ²³ atoms
0.2 mol -------------- x
x = (0.2 x 6.023 x 10 ²³) / 1
= 1.205 x 10²³ atoms
The answer is: <span>The principal idea here is how r they obtained:
for example: sodium u put equal molar concentrations of sodium (Na) and sodium ion (Na+) together in a beaker , then dip in this solution a platinium wire (zero potential) which is connected to a normal hydrogen electrode (electrode with zero potential) then u see the reading of the whole circuit
if it is negtive, this means negative potential which means that the reducing property predominates where Na(reducing agent) is oxidized and electrons r accumulated on the platinum which gives it negative charge
This means that (Na) is a reducing agent, its strength depends on the value of the potential obtained, and here the table can help you
If u want to know if it's strong red. agent, look for it in the table, see if it has higher reduction potentail (or lower oxidation potential, same idea) than most other substances then it is reducing agent
and vice versa
So if we look at ur examples, u will find that MnO4- is the very strong oxidizing agent (has highest oxidation potential) (lowest reduction potential)
H+ and H2 are both with zero potential, no redox properties
And lastly Na and Na+:
This u can know from ur knowledge in chemistry, that sodium is very rarely found in elemental form and always in the form of ion so u can deduce that Na is the very strong reducing agent
or u can see the value of its standard oxi or red potetial and deduce which is the predominating form of them.
I hope this helps</span>