1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains 0.110 M HNO2 and 0.170 M NaNO2. How many moles of HNO2 and Na
1 answer:
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Answer: 5 moles of can be produced from 5 mol NaHCO3 and 9 mol HCl.
Explanation:
The balanced chemical reaction is:
According to stoichiometry :
1 mole of use 1 mole of
Thus 5 moles of use=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of give = 1 mole of
Thus 5 moles of give =
of
5 moles of can be produced from 5 mol
and 9 mol HCl.
Answer:
In 8 drops (0.20 mL) of the solution there is 0.0060 g of starch.
Explanation:
A 3.0% (w/v) is a weight/volume percentage that means that there are 3.0 g of solute (starch) per 100 mL of solution. Then, in 0.20 mL of solution: