<span>Correct answer is:

But how to get there?
Let's start with simple explanation of what exactly is cellular respiration.
Cellular respiration is a multistage biochemical oxidation process of organic substances when prime product is energy (ATP - adenosine triphosphate) and other are released waste products. Cellular respiration takes place even if other metabolic processes are stopped, but cellular respiration may differ in particular organism groups.Some reactions during whole process of cellular respiration are similar in all types of living organisms.
Cellular respiration is prime indication of declining living processes.Only viruses which are on the edge of living organism and chemical particle are not performing cellular respiration.But to the point :P
In cellular respiration all substrates which are in the cell might be organic, but mostly we are using sugar oxidation - glucose in the presence of oxygen. Chemical formula of sugar looks like this:

Oxygen is just

so for now we have just part of the equation:

But what would be on the right hand side?
It's quite simple, remember equation of full combustion? If we want to burn something we need oxygen like in the equation, so the product of this equation would be carbon dioxide, water and of course energy (ATP).Carbon dioxide formula looks like this:

As a reminder water formula:

Full formula would look like that:

But still as you see this equation is unbalanced, after balancing it would like that:

At the end I would like to explain one more thing. Energy which has been released during this process is part of high-energy connection which might be used to perform chemical reactions in the cell or to move organism for example in muscles. We need to remember that production of ATP is not happening with 100% efficiency and part of this energy is released as heat.</span>
the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)
According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.
<h3>Definition of percent composition </h3>
The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.
To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.
<h3>Percentage Composition in this case</h3>
In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.
Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

<u><em>percentage composition of potassium iodide= 0.017%</em></u>
Finally, the percent composition of potassium iodide in this sample is 0.017%.
Learn more about percent composition:
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