You must burn 1.17 g C to obtain 2.21 L CO2 at
STP.
The balanced chemical equation is
C+02+ CO2.
Step 1. Convert litres of CO, to moles of CO2.
STP is 0 °C and 1 bar. At STP the volume of 1 mol
of an ideal gas is 22.71 L.
Moles of CO2= 2.21 L CO2 × (1 mol CO2/22.71 L
CO2) = 0.097 31 mol CO2
Step 2. Use the molar ratio of C:CO2 to convert
moles of CO to moles of C
Moles of C= 0.097 31mol CO2 × (1 mol C/1 mol
CO2) = 0.097 31mol C
Step 3. Use the molar mass of C to calculate the
mass of C
Mass of C= 0.097 31mol C × (12.01 g C/1 mol C) =
1.17 g C
It looks as if you are using the old (pre-1982)
definition of STP. That definition gives a value of
1.18 g C.
Answer:
0.03
Explanation:
22.8 g Ba(OH)2 (1 mol Ba (OH)2/ 171.34 g) = 0.133 mol Ba (OH)2
77.2 g H2O (1 mol H2O/18 g) = 4.29 mol H2O
X= molar fraction= mol Ba(OH)2/ mol total
X= 0.133/ (0.133+4.29) = 0.03
Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
- CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
Answer:
They are most likely solid
Explanation:
solid is a physical property
