The noble gas notation is the short or abbreviated form of the electron configuration.
It means that you use the symbol of the previous noble gas as part of the electron configuration of an element.
The gas noble previous to antimony is Kr, so you do not use Xe to write the electron configuration of Sb.
The gas noble previous to radium is Rn, so you do not use Xe to wirte the electron configuration of Ra.
The gas noble previous to uranium is Rn, so you do not use Xe to write the electron configuration of U.
The gas noble previous to cesium is Xe, so you use Xe to write the noble notation for Sb. This is it: Cs: [Xe] 6s.
Answer: cesium
The ga
Answer:
The energy released in the decay process = 18.63 keV
Explanation:
To solve this question, we have to calculate the binding energy of each isotope and then take the difference.
The mass of Tritium = 3.016049 amu.
So,the binding energy of Tritium = 3.016049 *931.494 MeV
= 2809.43155 MeV.
The mass of Helium 3 = 3.016029 amu.
So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV
= 2809.41292 MeV.
The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV
1 MeV = 1000keV.
Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.
So, the energy released in the decay process = 18.63 keV.
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Answer:
See explanation and image attached
Explanation:
When the carbocation is formed by the action of AlCl3 on the (CH3)3CCH2Cl, a primary carbocation is formed. The formation of the carbonation is followed by a 1,2-alkyl shift to give a tertiary carbocation which subsequently adds to the benzene ring as shown in the image attached to this answer.
Answer:
Follows are the explanation to this question:
Explanation:
When the drug is negatively charged, its negative electrolyte is annihilated to just the positive electrode. It is enticed, and it may not have a picture showing the electrode, however, We suppose that electrodes from either side of a skin slice. Its negative electrode will bypass or push thru the skin if in front of the counter terminal this becomes a red-positive electrode.
