Double replacement.
Ba(OH)2 + H2SO4 ====> 2H2O + BaSO4
Answer:
its either all animals or only mammals. the thing with all animals is that they do have life cycles but not the same as like say humans or elephants, some are complicated.
Explanation:
im sorry but it depends on what the diagram looks like for the problem
Answer with Explanation:
"Mass" and "weight" should never be used interchangeably with each other. Mass refers to the <u>total amount of matter</u><u> that can be measured in an object, </u>while weight refers to the<u> measure of the</u><u> force of gravity</u><u> that is acting on the object's mass.</u>
The mass of an object is<u> constant</u> (meaning, it doesn't change even if the object will be placed on another location) while the weight of an object relies on the <em>force of gravity.</em> So, this means that your mass on Earth and on the moon are identical, however, your weight on Earth and on the Moon are different. You will weigh lesser on the Moon because it has a lesser surface gravity than that of Earth.
So, this explains the answer.
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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