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Answered by : ❝ AǫᴜᴀWɪᴢ ❞
<span>As we know that
1 cu cm H2O = 1 mL H2O = 1g H2O
now
Heat of fusion of water = 79.8 cal/g
and
Heat of vaporization of water = 540 cal/g
Atomic weight of water : H=1 O=16 H2O=18
now by calculating and putting values
65.5gH2O x 79.8cal/gH2O x 1gH2O/540cal = 9.68g H2O (steam)
9.68gH2O x 1molH2O/18gH2O x 22.4LH2O/1molH2O = 12.0 L H2O
hope it helps</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+