In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following 
, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
Learn more:
The compound that has higher mass percent is Mg3N2 which is 72%
explanation
calculate the % mass of Mg in each compound
% composition of mg= molar mass of Mg/ total molar mass of the compound x100
in MgO = 24 /40 x100 = 60%
in Mg3N2 =72/100 x100 = 72 %
therefore the % composition in Mg3N2 is higher
Answer:
Mannose is a simple sugar or monosaccharide that is found as part of some polysaccharides in plants and in some animal glycoproteins. Galactose is converted to glucose in the liver to serve as fuel for cells in the body.
Explanation:
Answer:
The answer to your question is False.
Explanation:
Data
1 mole of NH₄NO₃
0.75 moles of N₂O
Percent yield = 25%
Chemical reaction
NH₄NO₃ ⇒ N₂O + 2H₂O
Process
1.- Determine the theoretical yield
1 mol NH₄NO₃ ------------- 1 mol of N₂O
2.- Calculate the percent yield
Percent yield = Actual yield / Theoretical yield x 100
-Substitution
Percent yield = 0.75 / 1 x 100
-Simplification
Percent yield = 0.75 x 100
-Result
Percent yield = 75%
Conclusion
False, the actual percent yield is 75%
No, actually adawadawada and awawawaw usually addawadadaw but also awawawa so it’s a possibility but very rare.
