The symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.
A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.
The actual electronic configuration of antimony is as follow:
[Kr] 4d10 5s2 5p3
B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.
The actual electronic configuration of Cesium is as follow:
[Xe] 6s1
Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.
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Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O
Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...
Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...
Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....
Do the calculation
Answer:
CH3 - CH2 - NH - C(CH3)3
Explanation:
The 1HNMR is the acronym for Proton nuclear magnetic resonance and it is used in chemistry or Chemical laboratory for the analysis and identification of compounds.
In order to be able to answer the question accurately you have to have the tables containing all the values for chemical shift. The Chemical shift is measured in ppm and it occur due to Resonance frequency variation.
From the table, a singlet at 1.15 ppm (9H) is - C(CH3)3.
A singlet at d 0.9 ppm (1H) shows the presence of a secondary amine group, that is -R2NH group.
A triplet at 1.10 ppm (3H) shows that we have;
CH3- CH2-
A quartet at 2.6 ppm (2H) shows that we have;
-CH2 - CH3.
Therefore, joining all together we have;
CH3 - CH2 - NH - C(CH3)3.
Kindly check attached file for the picture of the structure.
A serial dilution is the stepwise dilution of a substance in solution. Usually the dilution factor at each step is constant, resulting in a geometric progression of the concentration in a logarithmic fashion.
Answer:
0.4 moles of water produced by 6.25 g of oxygen.
Explanation:
Given data:
Mass of oxygen = 6.25 g
Moles of water produced = ?
Solution:
Chemical equation;
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 6.35 g/ 32 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of oxygen with water:
O₂ : H₂O
1 : 2
0.2 : 2×0.2 = 0.4 mol
0.4 moles of water produced by 6.25 g of oxygen.
