
Here we go ~
1 mole of
has 6.022 × 10²³ molecules of the given compound.
So, 0.78 mole of
will have ~


<span>Ethoxyethane; trifluoroborane; BF3.Et2O; Boron trifluoride ethyl ether; Boron trifluoride diethyl ether; Boron trifluoride-diethyl ether; Boron
</span>
Answer:

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+
Answer:
The reactants would appear at a higher energy state than the products.
Have a nice day!