sodium element is likely to have the highest thermal conductivity
Answer: 10 cm3
Explanation:
Mass of sample of steel (m) = 80 grams
density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?
Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:
density = mass / volume
p = m / v
make v the subject formula
v = m / p
v = 80 grams / 8 g/cm3
v = 10 cm3
Thus, the samples volume is 10 cm3
"<span>Changes of state are physical changes. They occur when matter absorbs or loses energy. Processes in which matter changes between liquid and solid states are freezing and </span>melting<span>. Processes in which matter changes between liquid and gaseous states are vaporization, evaporation, and condensation."</span>
[Kr]4d3
The configuration of Nb2+ is [Kr]4d3. Nb2+ also has an odd number of valence electrons.
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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