Enter a balanced equation for the reaction between aqueous lead(II) nitrite and aqueous potassium bromide to form solid lead(II)
1 answer:
Answer:
Explanation:
The charges are as follows:
since oxidation number of lead is 2 as indicated lead (II)nitrite. The number in parentheses indicates the oxidation number.
bears the charge -1.
As lead bears 2 positive charges so we need two nitrite ions to form lead (II) nitrite.
, it is a halogen and bears the charge -1.
, it is an alkali metal.
Since Bromide bears -1 charge and potassium bears +1 charge so we need only 1 ion of bromide and 1 ion of potassium to form potassium bromide.
Similarly, to form lead bromide we need 2 ions of bromide and 1 ion of lead, and to form potassium nitrite we need 1 nitrite ion and 1 potassium ion.
The final balanced equation is as follows, the various states of reactants and products are also written properly.
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Answer:
See explanation.
Explanation:
Hello,
In this case, we can show how the empirical formula is found by following the shown below procedure:
1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:
2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:
3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:
4. Compute the moles of oxygen by using its molar mass:
5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:
6. Search for the closest whole number (in this case multiply by 2):
Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:
Which is about three times in the molecular formula, for that reason, the actual formula is:
It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.
Best regards.