At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D
6_____18 record your data to be used in the following problem.
Answer:NaOH
Explanation:all are acids except NaOH
I am only in 6th grade i dont know this
Answer: 12.5 mg
Explanation:
The half-life of Iodine -123 is 13 hours.
Between 8 am on Monday and 10 am on Tuesday will be 26 hours.
After the first 13 hours, the Iodine should be left with half of its initial quantity
= 50/2
= 25 mg
There are now 13 hours remaining meaning that the Iodine will be half of the 25mg after that period.
= 25/2
= 12.5 mg.
12.5 mg remain at 10 am the next day.
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