Answer:
CHCL3= 11.73 mL
CHBr3= 8.268 mL
Explanation:
Let x be the mL of CHCl3 and y be the mL of CHBr3, then we have:
y+x = 20 mL
1.492x+2.89y=2.07* 20.0 mL
B.
its the only one with ozone in the reaction O₃
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Answer:
C. Hydrogen Ions.
Explanation:
The pH level of soils is measured based upon the acidity or basicity level in the soil. The pH scale usually measures between 0-14. 7 is the neutral pH level of soil and a pH level below 7 is considered acidic and above 7 is alkaline.
So, soil with a high amount of hydrogen ions (H+) will be considered acidic. As the amount of hydrogen ions increases in the soil, the pH level of the soil decreases, making it acidic.
Therefore, option C is correct.
