Applications of iron oxide nanoparticles include terabit magnetic storage devices, catalysis, sensors, superparamagnetic relaxometry (SPMR), and high-sensitivity biomolecular magnetic resonance imaging (MRI) for medical diagnosis and therapeutics.
Explanation:
The given data is as follows.
Concentration = 3.3 mol
Rate constant = 
Molecular weight = 55000 g/mol
Initial concentration (
) of aminoheptanoic is 3.3 kg/mol.
Hence, putting the given values into the above formula as follows.
= ![\frac{3.3}{2.74 \times 10^{3} kg/mol min} [X_{A}]^{0.1}_{0}](https://tex.z-dn.net/?f=%5Cfrac%7B3.3%7D%7B2.74%20%5Ctimes%2010%5E%7B3%7D%20kg%2Fmol%20min%7D%20%5BX_%7BA%7D%5D%5E%7B0.1%7D_%7B0%7D)
= 120.43 min
Thus, we can conclude that total time required to to form polyamide is 120.43 min.
It depends on the system if there are more moles of gas on the left and you increase the pressure the yield of a product will increase as the equilibrium moves to the side with the fewest gas molecules
if the right hand side has more moles of gas and you increase the pressure the yield will decrease as the equilibrium shifts to the left hand side where there are more gas moles
hope that helps
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.
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)
