A 0.7834 g sample of a primary standard KHP was titrated with 38.45 mL of NaOH solution to a phenolphthalein end-point. What is
1 answer:
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Explanation:
Step 1:
A good first step for a problem like this is to write down the chemical formula and balance it.
It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:
CH3COOH + NaOH ------> NaCH3COO + H2O
All of the constituents come out to a value of 1, conveniently.
Step 2:
Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:
M1 * V1 = M2 * V2
M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.
So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.
Our V1 is going to be the initial volume of the sample, which was 10.5 mL
Our V2 is going to be 19.13, which is the volume when we're finished.
It's clear that we don't know M2, so let's find it.
Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.
Using some algebra, we can see that we now have:
0.460 M * 0.0105 L = x M * 0.01913 L
Which goes to:
Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles =
Number of moles = = 0.003mole
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.
where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.
It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.
The oxidation state of Pb in is 2. So, moles deposited by Pb is as follows.
It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.
Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.