The question is incomplete, here is the complete question:
Consider the following reaction: 
In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?
<u>Answer:</u> The average rate of appearance of oxygen gas is 
<u>Explanation:</u>
We are given:
Moles of oxygen gas = 
Volume of solution = 0.480 L
Molarity is calculated by using the equation:

So, 
The given chemical reaction follows:

The average rate of the reaction for appearance of
is given as:
![\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%20of%20appearance%20of%20%7DO_2%3D%5Cfrac%7B%5CDelta%20%5BO_2%5D%7D%7B%5CDelta%20t%7D)
Or,

where,
= final concentration of oxygen gas = 0.0396 M
= initial concentration of oxygen gas = 0 M
= final time = 15.0 s
= initial time = 0 s
Putting values in above equation, we get:

Hence, the average rate of appearance of oxygen gas is 
Answer:
Dependent variable
Explanation:
In every experiment, there must be a dependent variable and an independent variable.
The independent variable is that variable that you have to manipulate in order to receive a response. E.g, in the experiment described in the question, the amount of sunlight a sunflower plant receives is the independent variable.
The dependent variable is the number of seeds produced. It is the response obtained by carefully manipulating the amount of sunlight
Answer:
Second Earth's Atmosphere
The second atmosphere, which was the first to remain with the earth, was created by volcanic outgassing and comet ice. There was a lot of water vapor, carbon dioxide, phosphorus, and methane in this atmosphere, but absolutely no oxygen.
Explanation:
The planet that orbits in a near circle is EARTH :D
As you move from left to right across a period, the number of protons in the nucleus increases. The electrons are thus attracted to the nucleus more<span> strongly, and the atomic radius is </span><span>smaller</span>