Explanation:
Expression for the kinetic energy is as follows.
K.E =
Now, total kinetic energy will be as follows.
K.E =
=
Since, this energy converts into electromagnetic radiation of wavelength 121.6 nm.
Relation between energy and photon is as follows.
Energy of photon =
=
=
v =
=
m/s
Thus, we can conclude that atoms were moving at a speed of
m/s before the collision.
Answer:
Both of the studies said that the mass of the atom is centered in the nucleus, which is positive, and there are electrons (negative particles) orbiting it. So, Rutheford and Nagaoka discovered that the atom can be divisible and it has an empty space.
But, in the model of Nagaoka, the nucleus was huge, and for Rutherford, the nucleus was really small, and the mass was concentrated. By his experiment with the gold sheets, the theory was appropriated. That's why Rutherford is credited with the discovery of the nucleus. Nagaoka was incorrect in his suppositions.
Acceleration can be described as changes in speed, changes in direction, or changes in both. Acceleration is a vector.
Answer:
6 moles of NaCl are needed to make 3.0 liters of a 2.0 M NaCl solution.
Explanation:
A 2M solution means that there are 2 moles of solute (NaCl in this case) in 1 liter of solution:
1 L solution-----2 moles of NaCl
3 L solution----x= (3 L solutionx2 moles of NaCl)/1 L solution= <em>6 moles of NaCl</em>
Answer : The concentration of
and
are
and
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 


Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of
and
are
and
respectively.