Answer:
I think it's B
Explanation:
I dont have much experience with the periodic table, but I just think its B.
Solution here,
Volume(V)=67.4 L
Pressure(P)=1 atm
Temperature(T)=(0+273)K=273K
Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1
No. of moles(n)=?
Now,
PV=nRT
or, 1×67.4=n×0.0821×273
or, 67.4=22.4n
or, n=67.4/22.4
or, n=3
therefore, required no. of mole is 3.
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:
1. CaCO3 + 2HCl → CaCl2 + H2O + CO2
2. C6H12O2 + 8O2 → 6CO2 + 6H2O
Explanation:
Answer:
we can do it again and again and again and again and again and again
