First, determine the amount of substance in moles by dividing the given amount in grams by the molar mass.
17.8 grams LiF x (1 mole / 25.94 grams LiF) = 0.686 moles LiF
Divide this amount of substance by the volume of the solution in liters.
molarity = 0.686 moles LiF / 0.915 L sol'n
The molarity is therefore, 0.75 M.
Answer:
Accuracy of values
Explanation:
They're correct due to margin error of ±0.5 during an experiment.
1 min 15 seconds is equivalent to 75 seconds.
However the other one was 75.5 seconds.
Using the margin error of ±0.5 during recording, they're both correct.
Note : Accuracy is the closeness of agreement between a measured value and a true or accepted value. Measurement error is the amount of inaccuracy.
Precision is a measure of how well a result can be determined (without reference to a theoretical or true value). It is the degree of consistency and agreement among independent measurements of the same quantity; also the reliability or reproducibility of the result.
The uncertainty estimate associated with a measurement should account for both the accuracy and precision of the measurement.
In this case, this is an argument of who was more accurate and judging by this standard, they're both accurate but not necessarily precise since their values are not exact with each other.
The equation is KE= 1.5 nRT
R = gas constant and it equal to 8.314 J/K- mol
T = Temperature
<span> Temperature for this problem is 125 Kelvin and n is equal to 1 mole.
Hope that helps u
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<span>*the ratio of moles of solute to liters of solution </span>
Answer:
The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas
Explanation:
There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.
These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.
For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.