When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal
1 answer:
Answer:
A is denser than B as it's volume for the same mass is smaller.
Explanation:
Hello.
In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:
Next, we compute the densities as shown below:
In such a way, A is denser is B as it's volume for the same mass is smaller.
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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: