Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
The answer is D because they would have to make quarries just below the surface to mine
Mineral ores much quicker (Quarries- a large hole/pit under ground used to get stone and minerals)
The problem you have written you almost have it solved. Take the moles that you have calculated and multiply that by the molecular weight to get the grams.
The STP problem:
use the moles you calculated along with 1 atm for Pressure, and 273 for the temperature and plug into the PV = nRT equation. (also use 0.0821 for R)
From there you can solve for the volume
Hope this helps!
Answer:
The ration of molar concentration is "2.5".
Explanation:
The given values are:
Average density of salt water,
= 
Net pressure,
= 
Increase in pressure,
= 
Now,
The under water pressure will be:
= 
= 
= 
= 
hence,
The ratio will be:
= 
or,
= 
= 
= 
Charge and uncharged particles