Model 1-How is a cell like a factory

Chemistry

1 answer:

anastassius [24]3 years ago

8 0

A cell is like a factory in many ways.

For one, the cell typically contains many organelles that have several varying functions. In a factory, you have many workers who perform different tasks, just as the organelles in the cell do.

Another reason is that cells contain a nucleus or the boss/brain of the cell. In factories, this could be considered the boss of the workplace that tells each and every worker, or organelle, what to do.

The selectively permeable membrane of a cell also resembles a factory as it only lets in workers or special guests. The cell's membrane only lets in specific materials, hence the selectively permeable membrane.

These are just a few ways as to how the cell is like a factory.

Hope this helps!

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Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.

Sophie [7]

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:

$y=-3.5 x + 25\\\\y=-0.52 x + 2$

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

$-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72$

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

$y=-3.5 (7.72) + 25\\\\y = 1.84$

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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