Al2(SO4)3 would be the formula.
Answer:
Explanation:
no. of moles of butane = 5.87g / 58.124 = 0.1009mol
since O2 is excess, limiting reagent is butane
from the equation, 2 moles of butane will reacted to produce water = 10 moles
therefore, 0.1009 moles of butane produced water = (0.1009 x 5) = 0.5050 moles
Weight of H2O formed = 0.5050 x 18.015 = 9.10 grams
HClO4(aq) ===> H+(aq) + ClO4-(aq)
If [HClO4] = 0.175 M, then [H+] = [ClO4-] = 0.175 M
[OH-] = 0.175 M = 1.75 x 10^-1 M
Kw = [H+][OH-} = 1.00 x 10^-14 [OH-] = 1.00x10^-4/1.75x10^-1 = 5.71 x 10^-14 M
Answer:
5.4 x 10²²molecules
Explanation:
Given parameters = 2L
Unknown; number of molecules
Condition of the water vapor is at standard Temperature and Pressure
Solution
To solve this problem, we simply find the number of moles the given volume would yield at STP using the relationship below:
Number of moles =
Number of moles = = 0.089mole
To find the number of molecules in the solved mole of the water vapor:
we use the avogadro's constant to evaluate.
1 mole of a water vapor = 6.02 x 10²³ molecules
0.089 mole of water vapor = 0.54 x 10²³molecules
Number of molecules of water is 5.4 x 10²²molecules