Answer:
P₂ = 300 atm
Explanation:
Given that,
Initial volume, V₁ = 600 L
Initial pressure, P₁ = 400 atm
We need to find the pressure if the volume is 800 L.
We know that the relation between pressure and volume is given by :

So, the new pressure is equal to 300 atm.
True i really don’t feel like giving an explanation but trust me it’s trye
Answer:
Of course it's C
Red planet
Explanation:
It is because the soil on Mars is rich of Fe (Iron).
That makes the soil look red.
Even on our planet we have such this places like hormuz island in Iran.
Answer:
is the oxidizing agent
Explanation:
An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.
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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