Answer:
1) 0.0025 mol/L.s.
2) 0.0025 mol/L.s.
Explanation:
<em>H₂ + Cl₂ → 2HCl.</em>
<em></em>
<em>The average reaction rate = - Δ[H₂]/Δt = - Δ[Cl₂]/Δt = 1/2 Δ[HCl]/Δt</em>
<em></em>
<em>1. Calculate the average reaction rate expressed in moles H₂ consumed per liter per second.</em>
<em></em>
The average reaction rate expressed in moles H₂ consumed per liter per second = - Δ[H₂]/Δt = - (0.02 M - 0.03 M)/(4.0 s) = 0.0025 mol/L.s.
<em>2. Calculate the average reaction rate expressed in moles CI₂ consumed per liter per second.</em>
<em></em>
The average reaction rate expressed in moles Cl₂ consumed per liter per second = - Δ[Cl₂]/Δt = - (0.04 M - 0.05 M)/(4.0 s) = 0.0025 mol/L.s.
Answer:
(a) (i) All the elements of a group have similar chemical properties because they have same no. of valence electrons in their outermost shell. (ii) All the elements of a period have different chemical properties because they have different no. of valence electrons in their atoms.
To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.
Let us determine the reaction for the titration below:
2NaOH +2H2SO4 = Na2SO4 +2H2O
So,
0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution
= 2.62 M H2SO4
The answer is the fourth option:
<span>2.62 M</span>
The given question is incomplete. The complete question is as follows.
Sodium sulfate is slowly added to a solution containing 0.0500 M 
and 0.0390 M 
. What will be the concentration of 
(aq) when 
begins to precipitate? What percentage of the can be separated from the Ag(aq) by selective precipitation?
Explanation:
The given reaction is as follows.
![[Ag^{+}]](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D)
= 0.0390 M
When 
precipitates then expression for 
will be as follows.
![K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]](https://tex.z-dn.net/?f=1.20%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%280.0390%29%5E%7B2%7D%20%5Ctimes%20%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![[SO^{2-}_{4}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 0.00788 M
Now, equation for dissociation of calcium sulfate is as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.00788)
![[Ca^{2+}]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D)
= 0.00625 M
Now, we will calculate the percentage of remaining in the solution as follows.
= 12.5%
And, the percentage of that can be separated is as follows.
100 - 12.5
= 87.5%
Thus, we can conclude that 87.5% will be the concentration of when begins to precipitate.
