Answer:
3 years
Explanation:
Given data:
Initial amount of sample = 160 Kg
Amount left after 12 years = 10 Kg
Half life = ?
Solution:
at time zero = 160 Kg
1st half life = 160/2 = 80 kg
2nd half life = 80/2 = 40 kg
3rd half life = 40 / 2 = 20 kg
4th half life = 20 / 2 = 10 kg
Half life:
HL = elapsed time / half life
12 years / 4 = 3 years
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
Well, a compound has a total charge of 0. So, it's electrically neutral. Since the X is 3+ and the Y is 3- they add to 0. Meaning no subscripts are necessary. Why don't you try a different combo?
Like:
A^3 and B^1-, to get a 3- charge you need 3xB^1- so the formula is AB3
Does this help?
Answer:
893 moles
Explanation:
An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are . 893 moles of Oxygen in 20.0 liters.
Answer: H+ ia helyuim
explinanation: Hope this helped!!
