From the pair of values given, the ratio of the volume to the temperature for the first data pair is, 0.0026 cm³/K
<h3>What is a Ratio?</h3>
A ratio is a comparison between two quantities showing how much one quantity differs from another.
From the data given, the first data pair is as follows ;
- Temperature = 276 K
- Volume = 0.72 cm³
Ratio of volume to temperature = volume / temperature
Ratio = 0.72 / 276
Ratio = 0.002608 cm³/K
The significant figure's rule for division is the least number of significant figures used in the operation.
The least significant figure of the values given is 2.
Therefore, the ratio of the volume to the temperature for the first data pair is 0.0026 cm³/K
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The case is acidic since calculated pH is below 7 that is 2.3
The hydroxide and hydrogen ion levels in a solution are the sources for calculating pOH and pH levels, respectively, and are the basis of whether a solution is acidic or basic
[H₃O+]=5.3×10⁻³M
To determine the hydroxide ion concentration, we use:
[OH−][H3O+]=1×10⁻¹⁴
We isolate the hydroxide ion concentration:
[OH−]=1×10−145.3×10⁻³M
Solving this, we obtain:
[OH−]=1.9×10−¹²M
We can determine whether the solution is acidic or basic by calculating the pH level:
pH=−log[H3O+]
We substitute in the hydronium ion concentration given:
pH=−Log(5.3×10−3)pH=2.3
Since the pH is below 7, the solution is acidic
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The answer i got was 2.55 M KOH
Hey there!
It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.
In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:
Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:
Thus, the density of the calcite sample will be:
This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL
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