The δs∘rxn for the reaction → will be -146 J/K.
Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.
Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit .
Entropy of the reaction can be calculated by the reaction.
Δ = 2 mol × × - 1 mol ×
Δ = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K
Δ = -146.8 J/K
Therefore, the δs∘rxn for the reaction → will be -146 J/K.
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Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
Whereas the moles of the salt are computed as shown below:
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
Whose equilibrium expression is:
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
Whereas x is:
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
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Answer:
Sodium bicarbonate
Explanation:
Sodium bicarbonate ( NaHCO₃ ) -
Sodium bicarbonate , according to the IUPAC nomenclature , its name is sodium hydrogen carbonate ,and in common terms also refereed to as baking soda .
It is a white crystalline solid , it is basic in nature .
<u>The cation and anion of this salt are the sodium ion ( Na⁺) and the anion bicarbonate anion (HCO³⁻) .</u>
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Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation: