Answer:
Explanation:
Fe⁺²(aq) + ClO₂(aq) → Fe⁺³(aq) + ClO₂⁻(aq)
Here oxidation number of Fe is increased from +2 to +3 , so Fe is oxidised .
The oxidation number of Cl is reduced from + 4 to +3 so Cl is reduced .
So ClO₂(aq) is oxidising agent and Fe⁺²(aq) is reducing agent .
Answer:
Atmosphere.
Explanation:
Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas.
The empirical formula is the simplest form of the formula expressed in the lowest ratio. In this case, we just have to divide each subscript by the greatest common factor. Hence.
a. CN
b. P2O5
c.N2O5
d.NaCl
e. C9H20
f. BH3
g.K2Cr2O7
h.AlB3
i.CH
j.SiCl4
Answer:
rats. that's all i know of Just about everything except the mother hen if they are natural hatch. Even when you incubate them there are threats. The healthy chicks will mob the weak ones, the older chicks (even by a day) will pick on the younger ones. Temperature extremes will threaten them as they need warm, humid conditions with gradual drops in surrounding temps in the brooder box. Early disease is sometimes a problem and all chicks should be started on medicated chick feed for the first few weeks to prevent several digestive diseases. Even the water dispenser can be a threat as newly hatched chicks will immerse themselves in an open water container so care should be taken to supply water in a self feeding covered dish.
Explanation:
Answer: The mass percentage of 
is 5.86%
Explanation:
To calculate the mass percentage of in the sample it is necessary to know the mass of the solute ( in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the 
precipitate. We can establish a relation between the mass of and using the stoichiometry of the compounds:
Since for every mole of Tl in there are two moles of Tl in , we have:
Using the molar mass of we have:
Finally, we can use the mass percentage formula:
