Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D)
a) moles of butane
![\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B5.2g%7D%7B58.12g%2Fmol%7D%3D0.09moles)
b) moles of oxygen
![\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B32.6g%7D%7B32g%2Fmol%7D%3D1.02moles)
![2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O](https://tex.z-dn.net/?f=2C_4H_%7B10%7D%2B13O_2%5Crightarrow%208CO_2%2B10H_2O)
According to stoichiometry :
2 moles of butane require 13 moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Thus 0.09 moles of butane will require =
of ![O_2](https://tex.z-dn.net/?f=O_2)
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
It changes rocks and minerals by water, ice, acids, salt, and changes in the temperature. Once the rock has been broken down a process named erosion happens, it transports bits of rocks and minerals away
Answer:
![\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B5.5%20%5Ctimes%2010%5E%7B22%7D%5Ctext%7B%20molecules%20of%20P%24_%7B2%7D%24O%7D_%7B5%7D%7D)
Explanation:
You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.
1. Moles of P₄O₁₀
![\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20P%24_%7B4%7D%24O%7D_%7B10%7D%20%3D%20%5Ctext%7B13%20g%20P%24_%7B4%7D%24O%7D_%7B10%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20P%24_%7B4%7D%24O%7D_%7B10%7D%7D%7B%5Ctext%7B283.89%20g%20P%24_%7B4%7D%24O%7D_%7B10%7D%7D%20%3D%20%5Ctext%7B0.0458%20mol%20P%24_%7B4%7D%24O%7D_%7B10%7D)
2. Moles of P₂O₅
P₄O₁₀ ⟶ 2P₂O₅
The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀
3. Molecules of P₂O₅
![\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}](https://tex.z-dn.net/?f=%5Ctext%7BNo.%20of%20molecules%7D%20%3D%20%5Ctext%7B0.0916%20mol%20P%24_%7B2%7D%24O%7D_%7B5%7D%20%5Ctimes%20%5Cdfrac%7B6.022%20%5Ctimes%2010%5E%7B23%7D%5Ctext%7B%20molecules%20P%24_%7B2%7D%24O%7D_%7B5%7D%7D%7B%5Ctext%7B1%20mol%20P%24_%7B2%7D%24O%7D_%7B5%7D%7D%5C%5C%5C%5C%3D%20%5Cmathbf%7B5.5%20%5Ctimes%2010%5E%7B22%7D%7D%5Ctextbf%7B%20molecules%20P%24_%7B2%7D%24O%7D_%7B5%7D%5C%5C%5Ctext%7BThere%20are%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B5.5%20%5Ctimes%2010%5E%7B22%7D%7D%5Ctextbf%7B%20molecules%20of%20P%24_%7B2%7D%24O%7D_%7B5%7D%7D%24%7D)
The empirical formula is the same as the molecular formula : C₁₀H₅O₂
<h3>Further explanation</h3>
Given
Molecular formula : C₁₀H₅O₂
Required
The empirical formula
Solution
The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.
The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
<em>(EF)n=MF
</em>
(EF)n = C₁₀H₅O₂
If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula
Answer and Explanation
The isomer picked is the N-Propylamine.
It has a lone pair of electron available on the electron rich Nitrogen and no formal charge.
Since it will be hard to draw the Lewis structure in this answer format, I'll attach a picture of the Lewis structure to this answer.
The lone pair of electron is shown by the two dots on the Nitrogen atom.