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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Answer:
your answer will be b) bonds are breaking
In a food chain, energy is passed through one link to another. When a herbivore eats only a certain fraction of the energy, (which comes from the food) it becomes new body mass; the rest of the energy is lost as waste or used up by the herbivore in order to carry out its life processes (ex. movement, digestion, reproduction). It doesn’t necessarily threaten the plants survival, there’s also a benefit. When a animals poops out the fruit (defecate) in another area those seeds get carried to new places with the help of a dab of fertilizer and a little bit of moisture. They also help supply nutrients when they die and decompose.
Answer:
True
Explanation:
Significant digits include non-zero digits (unless the zero is between two non-zero numbers)
In 12,785.000, there are 5 non-zero digits:
1 2 7 8 5
Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.
Explanation:
A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).
The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).
Here, the copper ions gain electrons and become solid copper.
