Answer:
pH = 9.03
Explanation:
The equilibrium of the NH₄Cl / NH₃ buffer in water is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Initial moles of both NH₃ and NH₄⁺ are:
0.100L ₓ (0.20 mol / L) = <em>0.0200 moles </em>
The NH₃ reacts with HCl producing NH₄⁺, thus:
NH₃ + HCl → NH₄⁺ + Cl⁻
<em>That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺</em>
Moles added of HCl were:
0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:
NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles
NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.
Using H-H equation for bases:
pOH = pKb + log [NH₄⁺] / [NH₃]
<em>Where pKb is -log Kb =</em><em> 4.745</em><em>.</em>
Replacing:
pOH = 4.745 + log 0.0250mol / 0.0150mol
pOH = 4.967
As pH = 14- pOH
<em>pH = 9.03</em>
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Ra, would have the lowest ionization energy. Remember ionization energy increases going up and to the right.
Answer:
![[CO]=[Cl_2]=0.01436M](https://tex.z-dn.net/?f=%5BCO%5D%3D%5BCl_2%5D%3D0.01436M)
![[COCl_2]=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.00064M)
Explanation:
Hello there!
In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:
![K=\frac{[CO][Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCO%5D%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Which can be written in terms of x, according to the ICE table:
Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:
![[COCl_2]=0.015M-0.01436M=0.00064M](https://tex.z-dn.net/?f=%5BCOCl_2%5D%3D0.015M-0.01436M%3D0.00064M)
Regards!
Answer:
1. As+H2=AsH3
2. draw a line under the problem with this:
A = 1 | A = 1
H = 2 | H = 3
3. We need everything to be equal (2x3 to get 6 then count the As on the product side to get the 2, then add 2 A on the reactant side):
A = 2 | A = 2
H = 6 | H = 6
4. 2As + 3H2 = 2AsH3
hope this helps
