When 1914.8C of electricity was passed through two electrolytic cells connected in series 0.6g of a metal M1 was deposited in one of the electrolyte, if 2.14g of another metal M2 was deposited at the other cathode and the sum of the quantity required to deposit 1mole of M1 and 1 mole of M2 is 289500C. Determine the charges on the ions of M1 and M2.
50.0 mL of a 0.200 M solution of the weak base N-ethylmorpholine (C6H13NO) is mixed with 8.00 mL of 1.00 M HCl and then diluted to a final volume of 100.0 mL with water, the result is a buffer with a pH of 7.00. Compute the Kb of N-ethylmorpholine.
mmols weak base = 50.0 x 0.200 = 10
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pK = pKw = 14 and solve for pKb and convert to Kb.
Note the typo in pKb.
mmols weak base = 50.0 x 0.200 = 10
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pKb = pKw = 14 and solve for pKb and convert to Kb.
Well, it seems like M1 and M2 have quite a shocking story to tell! Let's break it down with a little humor, shall we?
First, the electricity passed through the two electrolytic cells, and 0.6g of metal M1 was deposited in one of the electrolytes. Now, that's what I call a shocking transformation! M1 certainly knows how to make an electrifying appearance!
At the same time, 2.14g of metal M2 was deposited at the other cathode. Looks like M2 wanted to join the party and show off its own conductivity skills! Talk about a charged entrance!
Now, the sum of the quantity required to deposit 1 mole of M1 and 1 mole of M2 is given as 289500C. M1 and M2 really know how to add some serious wattage to the equation!
To determine the charges on the ions of M1 and M2, we'll have to use a little chemical wizardry. The charges of the ions can be calculated by dividing the total charge (289500C) by the number of moles of each metal deposited.
So, let's get mathematical here. Calculating the charge on the ions of M1 and M2 would be quite a shocking number indeed, but without knowing the number of moles deposited, it's like trying to juggle with rubber chickens. Hilarious, but not very helpful!
To solve this electrical mystery, we'll need more information about the number of moles of each metal deposited. Once we have those values, we can unleash the power of humor and chemistry to determine the charges on the ions of M1 and M2!
Keep the laughter flowing, my friend!
To determine the charges on the ions of M1 and M2, we need to use the Faraday's laws of electrolysis and perform calculations based on the given data and formula.
Faraday's laws state that the mass of a substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Let's break down the information given:
1. 1914.8C of electricity was passed through two electrolytic cells connected in series.
2. 0.6g of metal M1 was deposited in one of the electrolytes.
3. 2.14g of metal M2 was deposited at the other cathode.
4. The sum of the quantity required to deposit 1 mole of M1 and 1 mole of M2 is 289500C.
Using Faraday's laws, we can calculate the charges on the ions of M1 and M2. The charge (Q) is equal to the current (I) multiplied by time (t), so we can start by finding the value of current (I).
Given: Q = 1914.8C
Time (t) is not mentioned, so we'll assume it is 1 second.
Now, we can calculate the current (I):
I = Q / t = 1914.8C / 1s = 1914.8 A
Next, we'll find the amount of metal deposited (m) using the formula:
m = (Q / F) x M
where:
F is Faraday's constant, equal to 96500 C/mol (charge required to deposit 1 mole of metal).
M is the molar mass of the metal.
For metal M1:
m1 = (1914.8C / 96500 C/mol) x M1
Given that 0.6g of M1 was deposited, we can rearrange the equation to solve for M1:
M1 = (0.6g) x (96500 C/mol) / 1914.8C
M1 ≈ 30.62 g/mol
For metal M2:
m2 = (1914.8C / 96500 C/mol) x M2
Given that 2.14g of M2 was deposited, we can rearrange the equation to solve for M2:
M2 = (2.14g) x (96500 C/mol) / 1914.8C
M2 ≈ 107.24 g/mol
Finally, to determine the charges on the ions of M1 and M2, we look at their respective molar masses:
Charge for M1 = ±1 (assuming the ion carries a single charge)
Charge for M2 = ±2 (assuming the ion carries a double charge)
Therefore, the charge on the ions of metal M1 is ±1, and for metal M2 is ±2.
Please note that the charges are approximate values and are based on the given information and assumptions made during the calculations.