Answer:
B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.
Explanation:
Consider the mass of each particle. Express the masses in atomic mass units:
- Protons: approximately 1.007 amu each;
- Neutrons: approximately 1.009 amu each;
- Electrons: approximately 0.0005 amu each.
Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:
- Protons: +1 e;
- Neutrons: 0;
- Electrons: -1 e.
Calculate the charge-to-mass ratio for the three species:
- Protons: approximately
; - Neutrons: 0;
- Electrons: approximately
.
Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.
Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.
- The answer is shorter wavelength and equal speed.
That is, compared to ultraviolet light, an electromagnetic wave that has a higher frequency will also have shorter wavelength and equal speed.
This can be seen by the reaction given below:

h= Planck's constant
c=speed of the light
=frquency
=wavelength
So, higher is the frequency, lesser is the volume while speed remains constant as c is speed of light.
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7 is atomic symbol 9 is atomic density 8 is the atomic number
Answer: -105 kJ
Explanation:-
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%20B.E_%7BNH_3%7D%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN%5Cequiv%20N%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH-H%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BN-H%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20945%29%2B%283%5Ctimes%20432%29%5D-%5B%282%5Ctimes%203%5Ctimes%20391%29%5D)

Therefore, the enthalpy change for this reaction is, -105 kJ