Respuesta :
3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
Answer:
For 3: The total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]
For 4: The mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]
For 5: The equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]
Explanation:
- For 3:
To calculate the mass change of the reaction for given energy released, we use Einstein's equation:
[tex]E=\Delta mc^2[/tex]
E = Energy released = [tex]3.83\times 10^{-12}J[/tex]
[tex]\Delta m[/tex] = mass change = ?
c = speed of light = [tex]3\times 10^8m/s[/tex]
Putting values in above equation, we get:
[tex]3.83\times 10^{-12}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=4.255\times 10^3kg[/tex]
Hence, the total mass change of the reaction is [tex]4.255\times 10^{3}kg[/tex]
- For 4:
For the given isotopic representation: [tex]_{27}^{60}\textrm{Co}[/tex]
Atomic number = Number of protons = 27
Mass number = 60
Number of neutrons = Mass number - Atomic number = 60 - 27 = 33
To calculate the mass defect of the nucleus, we use the equation:
[tex]\Delta m=[(n_p\times m_p)+(n_n\times m_n)+]-M[/tex]
where,
[tex]n_p[/tex] = number of protons = 27
[tex]m_p[/tex] = mass of one proton = 1.00728 amu
[tex]n_n[/tex] = number of neutrons = 33
[tex]m_n[/tex] = mass of one neutron = 1.00867 amu
M = Nuclear mass number = 59.9338 amu
Putting values in above equation, we get:
[tex]\Delta m=[(27\times 1.00728)+(33\times 1.00867)]-[59.9338]\\\\\Delta m=0.54887amu[/tex]
Converting the value of amu into kilograms, we use the conversion factor:
[tex]1amu=1.66\times 10^{-27}kg[/tex]
So, [tex]0.54887amu=0.54887\times 1.66\times 10^{-27}kg=0.911\times 10^{-27}kg[/tex]
To calculate the equivalent energy, we use the equation:
[tex]E=\Delta mc^2[/tex]
E = Energy released = ?
[tex]\Delta m[/tex] = mass change = [tex]0.911\times 10^{-27}kg[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
Putting values in above equation, we get:
[tex]E=(0.911\times 10^{-27}kg)\times (3\times 10^8m/s)^2\\\\E=8.199\times 10^{-11}J[/tex]
Converting this into kilojoules, we use the conversion factor:
1 kJ = 1000 J
So, [tex]8.199\times 10^{-11}J=8.199\times 10^{-14}kJ[/tex]
Hence, the mass defect is [tex]0.911\times 10^{-27}kg[/tex] and energy equivalent to this mass is [tex]8.199\times 10^{-14}kJ[/tex]
- For 5:
For the given chemical reaction:
[tex]C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(l);\Delta H=-1418kJ/mol[/tex]
To calculate the equivalent mass of the reaction for given energy released, we use Einstein's equation:
[tex]E=\Delta mc^2[/tex]
E = Energy released = [tex]1418kJ=1418\times 10^3J[/tex]
[tex]\Delta m[/tex] = mass change = ?
c = speed of light = [tex]3\times 10^8m/s[/tex]
Putting values in above equation, we get:
[tex]1418\times 10^{3}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2\\\\\Delta m=1.5755\times 10^{-11}kg[/tex]
Hence, the equivalent mass of the reaction is [tex]1.5755\times 10^{-11}kg[/tex]
