Given 6 moles of CuCl2, how many moles of AlCl, were made? SHOW the math below
1 answer:
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Answer: The correct option is (c). The total pressure doubles.
Solution:
Initially, only 4 moles of oxygen gas were present in the flask.
(
) ( according to Dalton's law of partial pressure)
....(1)
= Total pressure when only oxygen gas was present.
Final total pressure when 4 moles of helium gas were added:
partial pressure of oxygen in the mixture :
Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.
= Total pressure of the mixture.
from (1)
On rearranging, we get:
The new total pressure will be twice of initial total pressure.
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.
So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then,
So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.