Answer:
= 72.73%
Explanation:
The percentage by mass of an element is given by;
% element = total mass of element in compounds/molar mass of compound × 100
The mass of oxygen in carbon dioxide = 32 g
Molar mass of CO2 = 44 g
Therefore;
% of O2 = 32/44 × 100%
<u>= 72.73%</u>
1.1 Moles / 0.5 Liters = 0.22 Molarity
The question is incomplete, complete question is :
A chemist must dilute 73.9 mL of 400 mM aqueous sodium carbonate solution until the concentration falls to 125 mM . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.
Answer:
The final volume of the solution will be 0.236 L.
Explanation:
Concentration of sodium carbonate solution before dilution =![M_1= 400 mM](https://tex.z-dn.net/?f=M_1%3D%20400%20mM)
Volume of sodium carbonate solution before dilution = ![V_1=73.9 mL](https://tex.z-dn.net/?f=V_1%3D73.9%20mL)
Concentration of sodium carbonate solution after dilution =![M_2= 125 mM](https://tex.z-dn.net/?f=M_2%3D%20125%20mM)
Volume of sodium carbonate solution after dilution = ![V_2=?](https://tex.z-dn.net/?f=V_2%3D%3F)
Dilution equation is given by:
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
![V_2=\frac{M_1V_1}{M_2}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BM_1V_1%7D%7BM_2%7D)
![V_2=\frac{400 mM\times 73.9 mL}{125 mM}= 236.48 mL\approx 236 mL](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7B400%20mM%5Ctimes%2073.9%20mL%7D%7B125%20mM%7D%3D%20236.48%20mL%5Capprox%20236%20mL)
1 mL = 0.001 L
236 mL = 0.236 L
The final volume of the solution will be 0.236 L.
All acids only conduct electric current when they are dissolved in water, because when they are in an aqueous medium, they undergo ionization, i.e. release <span>ions.</span><span>
HCl --> H</span>⁺ + Cl⁻
The bases also conduct electric current in solution, because the Ionic dissociation sufferers (release the existing ions in the formula) and molecular ionization, suffer the reacting with water and releasing ions.
NaOH ---> Na⁺ + OH⁻
hope this helps!