Answer:
The answer to your question is 0.54M
Explanation:
Data
Final concentration = ?
Concentration 1 = 0.850 M
Volume 1 = 249 ml = 0.249 l
Concentration 2 = 0.420 M
Volume 2 = 0.667 M
Process
1.- Calculate the number of moles in both solutions
Number of moles 1 = Molarity 1 x Volume 1
= 0.850 x 0.249
= 0.212
Number of moles 2 = Molarity 2 x Volume 2
= 0.420 x 0.667
= 0.280
Total number of moles = 0.212 + 0.280
= 0.492
2.-Calculate the final volume
Final volume = Volume 1 + Volume 2
Final volume = 0.249 + 0.667
= 0.916 l
3.- Calculate Molarity
Molarity = 0.492 / 0.916
Molarity = 0.54
3.4 molecules is in 127 grams of iodine
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1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0
Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf
[Ca(s)] were 0 because these are elements in their standard states.
</span>HOPE THIS HELPS ;)
