The uncertainties of the delta measurements and the uncertainty of the atomic weight derivedfrom the best measurement of isotopic abundances constrain the number of significant figures in theatomic-weight values of the upper and lower bounds. For carbon, the fifth digit after the decimal pointis uncertain because of the uncertainty value of 0.000 027. Therefore, the number of significant digitsin the atomic-weight value is reduced to four figures after the decimal point. The Commission may rec-ommend additional conservatism and reduce the number of significant figures further. For the lowerbound of carbon, 12.009 635 is truncated to 12.0096. For an upper bound, the trailing digit is increasedto ensure the atomic-weight interval encompasses the atomic-weight values of all normal materials. Inthe case of carbon, the upper bound is adjusted from 12.011 532 to 12.0116 to express four digits afterthe decimal point. The lower and upper bounds are evaluated so that the number of significant digits ineach is identical. If a value ends with a zero, it may need to be included in the value to express therequired number of digits. The following are examples of lower and upper atomic-weight bounds foroxygen that could be published by the Commission in its various tables.
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Extracted from" Atomic weights of the elements 2009 (IUPAC Technical Report)"
Answer: I think it’s the first one
This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
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