Answer:
0.962 atm.
97.4 kPa.
731 torr.
14.1 psi.
97,434.6 Pa.
Explanation:
Hello.
In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:
- Atmospheres: 1 atm = 760 mmHg:
- Kilopascals:: 101.3 kPa = 760 mmHg:
- Torrs: 760 torr = 760 mmHg:
- Pounds per square inch: 14.69 psi = 760 mmHg:
- Pascals: 101300 Pa = 760 mmHg:
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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.