Answer:
We know that
ħf = ф + Ekmax
where
ħ = planks constant = 6.626x10^-34 J s
f = frequency of incident light = 1.3x10^15 /s (1 Hz =
1/s)
ф = work function of the cesium = 2.14 eV
Ekmax = max kinetic energy of the emmitted electron.
We distinguish that:
1 eV = 1.602x10^-19 J
So:
2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J
So,
Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J
= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J
Answer:
5.19x10^-19 J
Kinetic energy:
In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.
Answer details:
Subject: Chemistry
Level: College
Keywords:
• Energy
• Kinetic energy
• Kinetic energy of emitted electrons
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Answer:
All other units are gotten from them
Answer:
The pH value of the mixture will be 7.00
Explanation:
Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,
![pH=pK_{a} + log(\frac{[Base]}{[Acid]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%29)
According to the given conditions, the equation will become as follow
![pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BNa_%7B2%7DHPO_%7B4%7D%20%5D%7D%7B%5BNaH_%7B2%7DPO_%7B4%7D%5D%7D%29)
The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.
Placing all the given data we obtain,
Answer:
the answer is 0 amu I hope it helps
