Answer:
1.14 atm and 1.139 mol
Explanation:
The <em>total pressure</em> of the container is equal to the <u>sum of the partial pressure of the three gasses</u>:
- P = Poxygen + Pnitrogen + Pcarbon dioxide
- 2.50 atm = 0.52 + 0.84 + Pcarbon dioxide
Now we <u>solve for the pressure of carbon dioxide</u>:
- Pcarbon dioxide = 1.14 atm
To c<u>alculate the number of CO₂ moles </u>we use <em>PV=nRT</em>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 32 °C ⇒ 32 + 273.16 = 305.16 K
1.14 atm * 25.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305.16 K
The element with the lowest ionization energy is CESIUM, CS.
Ionization energy is the energy required to remove the most loosely bound electron in an atom of an element. The higher the number of shells in an atom of an element, the lower the ionization energy that will be required to remove the valence electron.
I think it might be D or B
And my other two might be A or C
