me
pls work list for you and your business and I can say poop and you will be a moon
207 is the mass number. 82 would be the atomic number
Answer:
![\delta H_{rxn} = -66.0 \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20-66.0%20%20%5C%20kJ%2Fmole)
Explanation:
Given that:
![3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)](https://tex.z-dn.net/?f=3FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%5Cto%202Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%20-47.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%281%29%20%20%5C%5C%20%5C%5C%20%5C%5C%20Fe_2O_3_%7B%28s%29%7D%20%2B3CO_%7B%28g%29%7D%20%5Cto%202FE_%7B%28s%29%7D%20%2B%203CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-25.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%282%29%20%20%5C%5C%20%5C%5C%20%5C%5C%20Fe_3O_4_%7B%28s%29%7D%20%2B%20CO_%7B%28g%29%7D%20%5Cto%203FeO_%7B%28s%29%7D%20%2B%20CO_%7B2%28g%29%7D%20%5C%20%5Cdelta%20H%20%3D%2019.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%283%29)
From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:
![3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)](https://tex.z-dn.net/?f=3FeO_%7B%28s%29%7D%20%2B%20CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%20Fe_3O_4_%7B%28s%29%7D%20%2B%20CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-19.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%284%29)
Multiplying (2) with equation (4) ; we have:
![6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)](https://tex.z-dn.net/?f=6FeO_%7B%28s%29%7D%20%2B%202CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%202Fe_3O_4_%7B%28s%29%7D%20%2B%202CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-38.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%285%29)
From equation (1) ; multiplying (-1) with equation (1); we have:
![2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)](https://tex.z-dn.net/?f=2Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5Cto%20%20%20%20%203FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%20%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%2047.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%286%29)
From equation (2); multiplying (3) with equation (2); we have:
![3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)](https://tex.z-dn.net/?f=3%20Fe_2O_3_%7B%28s%29%7D%20%2B9CO_%7B%28g%29%7D%20%5Cto%206FE_%7B%28s%29%7D%20%2B%209CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-75.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%287%29)
Now; Adding up equation (5), (6) & (7) ; we get:
![6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)](https://tex.z-dn.net/?f=6FeO_%7B%28s%29%7D%20%2B%202CO_%7B2%28g%29%7D%20%20%20%20%5Cto%20%20%20%202Fe_3O_4_%7B%28s%29%7D%20%2B%202CO_%7B%28g%29%7D%20%20%20%5C%20%5Cdelta%20H%20%3D%20-38.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%285%29)
![2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)](https://tex.z-dn.net/?f=2Fe_3O_4_%7B%28s%29%7D%20%2BCO_%7B2%28g%29%7D%20%5Cto%20%20%20%20%203FeO_3_%7B%28s%29%7D%2BCO_%7B%28g%29%7D%20%20%20%5C%20%20%5C%20%5Cdelta%20H%20%3D%2047.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%286%29)
![3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)](https://tex.z-dn.net/?f=3%20Fe_2O_3_%7B%28s%29%7D%20%2B9CO_%7B%28g%29%7D%20%5Cto%206FE_%7B%28s%29%7D%20%2B%209CO_%7B2%28g%29%7D%20%20%5C%20%5C%20%5Cdelta%20H%20%3D%20-75.0%20%5C%20kJ%2Fmole%20%20--%20equation%20%287%29)
<u> </u>
![FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole](https://tex.z-dn.net/?f=FeO%20%20%5C%20%5C%20%5C%20%2B%20%20%5C%20%5C%20%5C%20CO%20%20%20%5C%20%5C%20%20%5Cto%20%20%20%5C%20%5C%20%5C%20%5C%20Fe_%7B%28s%29%7D%20%2B%20%5C%20%5C%20CO_%7B2%28g%29%7D%20%5C%20%5C%20%5C%20%20%5Cdelta%20H%20%3D%20-%2066.0%20%5C%20kJ%2Fmole)
<u> </u>
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(According to Hess Law)
![\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20%28-38.0%20%2B%20%2047.0%20%2B%20%28-75.0%29%29%20%5C%20kJ%2Fmole)
![\delta H_{rxn} = -66.0 \ kJ/mole](https://tex.z-dn.net/?f=%5Cdelta%20H_%7Brxn%7D%20%3D%20-66.0%20%20%5C%20kJ%2Fmole)
Answer:
A
Explanation:
molarity=moles of solute/liter of solution
molarity=0.26/0.3
molarity=0.87molar
Answer:
The atom must lose its three extra electrons to make the atom over all neutral.
Explanation:
The three subatomic particles construct an atom electron, proton and neutron. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons.
If an atom have -3 charge it means three more electrons are added. In order to make the atom overall neutral three more electrons must be removed so that negative and positive charge becomes equal and cancel the effect of each other and make the atom neutral.
Electron:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol= e⁻
Mass= 9.10938356×10⁻³¹ Kg
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
Proton and neutron:
While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.
Symbol of proton= P⁺
Symbol of neutron= n⁰
Mass of proton=1.672623×10⁻²⁷ Kg
Mass of neutron=1.674929×10⁻²⁷ Kg