How many moles are in 56.7 L of CO2
2 answers:
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
Finally, the resulting molarity in 30.8 mL (0.0308 L):
Regards.