Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
Hi there!
Zinc: Is qualitative
Chlorine: is quantitative
Gallium: is neither
Nitrogen: is quantitative
Aluminum: is quantitative
If you need an explanation, please let me know !
Hope this helps and have a good day :) !
~Angel
Hello, I Am BrotherEye
Answer:
As the atmospheric pressure decreases with increase in altitude, when a sealed bag of chips is taken to higher altitude then the pressure of the gases inside the bag become greater than the outer atmospheric pressure and apply pressure on the side covering of the chips bag due to which the chips bag expand.
Explanation:
Atmospheric pressure decreases with increasing altitude. Let's assume you tie up a bag at sea level altitude (with little air inside). That air inside the bag is at 1atm as long as you're at sea level.
Now you start going up, your altitude increases and atmospheric pressure decreases (remember what I said in the beginning about atmospheric pressure decreasing with increasing altitude?).
Let's say you reach an altitude where the atmospheric pressure is 0.5atm, the force acting inside the bag which is pushing it outward is now greater than the force acting on it which is pushing it inward (1atm>0.5atm). As a result, the air inside the bag will push the sides of the bag outward, thus inflating it to a point where the pressure inside the bag becomes 0.5atm or the bag stops stretching.
For the same reason, if you bring the same bag down and take it too deep into the earth (at an altitude lower than sea level) or if you inflate the bag at a very high altitude and bring it down to sea level, it will deflate.
Best Of Luck
~
BrotherEye
Answer : The initial concentration of HI and concentration of 
at equilibrium is, 0.27 M and 0.386 M respectively.
Solution : Given,
Initial concentration of 
and 
= 0.11 M
Concentration of and at equilibrium = 0.052 M
Let the initial concentration of HI be, C
The given equilibrium reaction is,
Initially 0.11 0.11 C
At equilibrium (0.11-x) (0.11-x) (C+2x)
As we are given that:
Concentration of and at equilibrium = 0.052 M = (0.11-x)
0.11 - x = 0.052
x = 0.11 - 0.052
x = 0.058 M
The expression of 
will be,
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
By solving the terms, we get:
C = 0.27 M
Thus, initial concentration of HI = C = 0.27 M
Thus, the concentration of at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M
So visualize soil mixed with water. To separate it, you can use a sieve to filter the water.
