The false is A
as C-14 is an unstable isotope that is radioactive and is present in all living things i can be used to carbon date things
hope that helps
100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)
Hope this helps~
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.
Answer: 6 moles
Take a look at the balanced chemical equation for this synthesis reaction
N 2(g] + 3 H 2(g] → 2 NH 3(g]
Notice that you have a 1:3 mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.
So, if you have 2 moles of nitrogen taking part in the reaction, you will need
2 moles N 2 ⋅ 3 moles H 2 /1 mole N 2 = 6 moles H 2