Answer:
See explanation
Explanation:
Substitution implies that one thing is repalced by the other.
Substitution of one ion by another largely depends on the sizes of the ions in question.
The ionic radius of Mn3+ is about 0.65 Angstroms while that of In3+ is about 0.94 Angstroms. On the other hand, the ionic radius of Y3+ is about 1.032 Angstroms.
It is clear that the sizes of Mn3+ and In3+ are closer to each other hence Mn3+ can substitute for In3+ easily than Y3+ in a lattice site.
The larger size of Y3+ explains why it is not easily replaced by Mn3+.
<u>Answer:</u> The heat of formation of oleic acid is -94.12 kJ/mol
<u>Explanation:</u>
We are given:
Heat of combustion of oleic acid = 
The chemical equation for the combustion of oleic acid follows:
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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(18\times \Delta H^o_f_{(CO_2(g))})+(17\times \Delta H^o_f_{(H_2O)})]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%2818%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%2817%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_%7B18%7DH_%7B34%7DO_2%28l%29%29%7D%29%2B%28%5Cfrac%7B51%7D%7B2%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-1.11\times 10^4=[(18\times (-393.51))+(17\times (-241.82))]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times 0)]\\\\\Delta H^o_f_{(C_{18}H_{34}O_2(l))}=-94.12kJ/mol](https://tex.z-dn.net/?f=-1.11%5Ctimes%2010%5E4%3D%5B%2818%5Ctimes%20%28-393.51%29%29%2B%2817%5Ctimes%20%28-241.82%29%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_%7B18%7DH_%7B34%7DO_2%28l%29%29%7D%29%2B%28%5Cfrac%7B51%7D%7B2%7D%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28C_%7B18%7DH_%7B34%7DO_2%28l%29%29%7D%3D-94.12kJ%2Fmol)
Hence, the heat of formation of oleic acid is -94.12 kJ/mol
Answer:
1.39 Atoms
Explanation:
When converting from moles to atoms, multiply the number of moles by Avogadro's number
Multiply 2.30 moles by Avogadro's number (6.022*10^23)
= 1.3851 Atoms
=1.39 Atoms (sigfig)
Answer:
Strontium is smaller
Strontium has the higher ionization energy
Strontium has more valence electrons
Explanation:
It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table
While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)
Since they are in the same period, periodic trends would be useful in evaluating their properties
In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size
Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius
Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has
In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case
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