Answer:
Isotopes of an element are atoms of the same element that have different number of neutrons.
Add water to mixture to dissolve sugar then filter. sugared water will be the filtrate while the salt & iron fillings will be the residue. use magnet to attract iron fillings and remaining will be salt.
Answer;
The above statement is true
upon heating a copper sample will expand, leading to a lower density
Explanation;
-The density of solids decreased with increase in temperature and vice versa. The increase in temperature causes the volume of the solid to increase which as a result decreases the density as Density=Mass/Volume. The temperature of a body is the average kinetic energy of the molecules present in it.
In other words; The temperature of a body is the average kinetic energy of the molecules present in it. Therefore; when heat is supplied ( or temperature is increased) the average kinetic energy increases which increases the volume and thus density decreases.
This
electronic transition would result in the emission of a photon with the highest
energy:
4p
– 2s
<span>This
can be the same with the emission of 4f to 2s which would emit energy in the
visible region. The energy in the visible region would emit more energy than in
the infrared region which makes this emission to have the highest energy.</span>
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
