Hello, 3Coli Here!
Your Answer is Here:
I would recheck, redo, and revise my answers, if necessary. Also, I can look at the lesson again, if I forgot something. So that's what I would do.
Hopefully, this helps!
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Answer: Volume of the balloon at this altitude is 46.9 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 737 mm Hg
= final pressure of gas = 385 mm Hg
= initial volume of gas = 28.6 L
= final volume of gas = ?
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get:
Thus the volume of the balloon at this altitude is 46.9 L
Answer:
A
Explanation:
You can eliminate C because you need to consult a periodic table to determine the mass (not the weight).
There are other answers you could use. The wording is awfully sloppy. D is an example of that. The elements part is true, the atoms part is not really true in all cases. For example NaCl is table salt and both the elements are ions, not atoms. Na is plus ion and Cl is minus one. B is wrong for the same reason.
I think in the end I would choose A. I think this is a very sly question. It is not really testing chemistry. It is testing how well you know your definitions.
<h2><u>
PLEASE MARK BRAINLIEST!</u></h2>
In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.
What is the concentration of H+ ions at a pH = 2?
<h3>
0.01 mol/L </h3>
What is the concentration of OH– ions at a pH = 2?
<h3>
0.000000000001 mol/L
</h3>
What is the ratio of H+ ions to OH– ions at a pH = 2?
<h3>
10,000,000,000 : 1</h3>
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Those are your correct answers on edg2020!
I LITERALLY spent 40 MINUTES trying to figure out this question, so please, use my VERY CORRECT answers!
<em>I hope this helps!</em>