Answer:
2s²2p⁴
Explanation:
Oxygen is an element on the periodic table with a total of 8 electrons. It's electronic configuration is given as 2,6.
Using the orbital notation we write as 1s²2s²2p⁴
Also, the valence electrons are the electrons in the outermost shell of an atom. These electrons mostly determine the chemical properties of an atom.
Oxygen has a total of 6 electrons in its outermost shell and it is given as 2s²2p⁴
Answer:
17.57kg of
and its percentage yield is 81.0%
Explanation:
Through the reaction you can get the theoretical amount of
that must be produced.

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage
=81.0%
Answer:
The correct answer is 0.300 * 10^23 ions.
Explanation:
Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.
The moles of zinc chloride (ZnCl2) is,
= mass of zinc + 2 mass of chlorine
= 65.38 + 2 (35.45)
=65.38 + 70.90
= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)
Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine
Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8
= 3.537 g of chlorine
70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537
= 0.300 * 10^23 ions of chlorine.
Answer:
a) find attached image 1
b) find attached image 2
Explanation :
The more stable radical is formed by a reaction with smaller bond dissociation energy.
since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.
<u>Answer:</u> The enthalpy of the reaction for the production of
is coming out to be -74.9 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CH_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-74.9%29%29%5D-%5B1%5Ctimes%200%29%2B%282%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-74.9kJ)
Hence, the enthalpy of the reaction for the production of
is coming out to be -74.9 kJ