I think it’s C atomic radius and numbers of unshielded protons
Answer:
1.28 mol
Explanation:
mole = mass/molar mass
n = v/v/cm³
mass = 0. 075g
v = 1dm³ =1000cm³
n= m/MV=0.075/58.44(1000)
n =1.28 mol
The answer is: D.unstable nuclei emitting high-energy particles as they formed more stable compositions.
Those high-energy particles are alpha particles
, beta particles
, gamma radiation.
For example, the decay chain of ²³⁸U is called the uranium series.
Decay start with U-238 and ends with Pb-206. There are several alpha and beta minus decays.
Antoine Henri Becquerel (1852 – 1908) was a French physicist and the first person to discover evidence of radioactivity.
Becquerel wrapped fluorescing crystal (uranium salt potassium uranyl sulfate) in a cloth, along with the photographic plate and a copper Maltese cross.
Several days later, he discovered that a image of the cross appeared on the plate.
The uranium salt was emitting radiation.
Because of this discovery, Becquerel won a Nobel Prize for Physics in 1903, which he shared with Marie Curie and Pierre Curie.
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>
<u />
(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)