C) 3 Moles
This is because of the molecular structure shown in the image
Answer:
After the transfer the pressure inside the 20 L vessel is 0.6 atm.
Explanation:
Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

Therefore, for this problem the step by step explanation is:

Clearing P2 and replacing

Answer ( 3 ) :
<span>conversion of matter to energy .
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hope this helps!
There you go , This is my work i have did on that subject area
B) 2
You would first balance GO4 by adding a coefficient of 3 in front of DnGO4 in the reactants. Then you’d balance the 3DnGO4 by adding a coefficient of 3 in front of Dn in the products. Finally you’d balance Eg by adding a coefficient of 2 in front of Eg to balance with the Eg2(GO4)3 in the products.