Answer:
2 moles
Explanation:
The following were obtained from the question:
Molarity = 0.25 M
Volume = 8L
Mole =?
Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:
Molarity = mole of solute/Volume of solution.
With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:
Molarity = mole of solute/Volume of solution.
0.25 = mole of MgCl2 /8
Cross multiply to express in linear form
Mole of MgCl2 = 0.25 x 8
Mole of MgCl2 = 2 moles
Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution
Answer is: ph value of pyridine solution is 9.1.
Chemical
reaction: C₅H₅N +
H₂O → C₅H₅NH⁺ + OH⁻.<span>
c(pyridine - C</span>₅H₅N)
= 0.115M.<span>
Kb(C</span>₅H₅N)
= 1.4·10⁻⁹.
[C₅H₅NH⁺] = [OH⁻] = x; equilibrium concentration.<span>
[</span>C₅H₅N] =
0.115 M - x.
Kb = [C₅H₅NH⁺] · [OH⁻] / [C₅H₅N].
1.4·10⁻⁹ = x² / (0.115 M -x)
Solve quadratic equation: x = [OH⁻] = 0.0000127 M.<span>
pOH = -log(0.0000127 M) = 4.9</span>
<span>pH = 14 - 4.9 = 9.1.</span>
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature 
= 
= rate constant at temperature 
= 
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)
The activation energy Ea for this reaction is 22689.8 J/mol
Answer:
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Explanation:
