Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
The K sp Value is 
Explanation:
From the question we are told that
The of 
is = 122.5 g/ mol
The mass of dissolved is 
The volume of solution is 
The number of moles of is mathematically evaluated as
Substituting values
Generally concentration is mathematically represented as
For
The dissociation reaction of is
The solubility product constant is mathematically represented as
Since there is no ionic reactant we have
![K_{sp} = [k^+] [ClO_3^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Bk%5E%2B%5D%20%5BClO_3%5E-%5D)
Answer:
Ionization energy
Electronegativity
Explanation:
-due to its smaller ionic radius....the electron in the outter most shell tends to expierence a stronger nuclear attraction...which makes it harder to remove the electron from the sodium atom
-Rubidium has lesser ionization energy because its (i) affected by its larger ionic radius which tends to lessen the nuclear attraction ...hence making it easier to remove the electron...(ii)and also by the screening effect done by the inner shells, which also tends to lessen the nuclear attraction.
Sodium has a higher electronegativity than rubidium;
Electronegativity is the charge density of electrons in an atom...in which its high when the atomic radius is smaller...
So hence due to the sodium atomic radius being smaller...it tends to have a higher charge density than rubidium....which then gives it a higher electronegativity value
Hybridization in ozone, O3......
<span>...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O </span>
<span>../ </span>
<span>O .................. 3 lone pairs on terminal O </span>
<span>I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. </span>
<span>The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. </span>
<span>If the electron pair geometry is linear, the hybridization is sp. </span>
<span>If the electron pair geometry is trigonal planar, the hybridization is sp2. </span>
<span>If the electron pair geometry is tetrahedral, the hybridization is sp3. </span>
<span>The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.</span>
Acid more H3O+ than OH-
Base less H3O+ than OH-
