Ans: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Answer : The volume of 
produced at standard conditions of temperature and pressure is 0.2422 L
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of 
gas = (740-22.4) torr = 717.6 torr
= final pressure of gas at STP= 760 torr
= initial volume of gas = 280 mL
= final volume of gas at STP = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Therefore, the volume of produced at standard conditions of temperature and pressure is 0.2422 L
Answer:
There are more nutrients available in estuaries. There are more nutrients available in estuaries.
Explanation:
