Using the accepted density of CO2 at STP, we get the expected volume of the gas.
0.10 g / (1.96g/L) = 0.051 g
Calculating for the percent error:
%error =( |0.051 - 0.056| / 0.051 ) (100) = 9.80%
The percent error is 9.80%.
Answer:
Sodium oxide is the product
Explanation:
4Na+O2->2Na2O
I hope this will help you
Answer:
K(eq) = 15 (2 sig. figs)
Explanation:
Rxn: CO(g) + 2H₂(g) ⇄ CH₃OH(l)
C(eq): 0.150M 0.360M 0.282M
Keq = [CH₃OH(l)]/[CO(g)][H₂(g)]
= (0.282M)/(0.150M)(0.360M)²
= 14.50617284 (calc. ans.)
= 15 (2 sig. figs.)
