Answer:
0.07906687 amu
Explanation:
For Boron ₅B¹¹, the number of protons is 5 and the mass is 11. The mass is the number of protons plus the number of neutrons, so:
neutrons = 11 - 5 = 6
The mass of an atom is concentrated in the nucleus, so it is the mass of the protons + the mass of the neutrons. The mass of 1 proton is 1.00727647 amu/proton, and the mass of 1 neutron: 1.00866492 amu/neutron, so for the element given the theoretical mass (mt) is:
mt = 5* 1.00727647 amu/proton + 6*1.00866492 amu/neutron
mt = 11.08837187 amu
The mass defect (md) is the theorical mass less the real mass:
md = 11.08837187 - 11.009305
md = 0.07906687 amu
<u>Answer:</u> The enthalpy of the reaction for the production of 
is coming out to be -74.9 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of is coming out to be -74.9 kJ
The volumeof this sphere must be less than 0.7228
Answer:
The answer is: A
Explanation:
The wave-like contractions are to be considered a physical change as they do not take part in the breaking-down of the food. The peptidase is a chemical change because it is used to chemically break down the food to retrieve nutrients and such.
