Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
Answer:
3.43×10¹ mol
Explanation:
Given data:
Initial number of moles = 12.4 mol
Initial volume = 122.8 L
Final number of moles = ?
Final volume = 339.2 L
Solution:
The number of moles and volume are directly proportional to each other at same temperature and pressure.
V₁/n₁ = V₂/n₂
122.8 L/ 12.4 mol = 339.2 L / n₂
n₂ = 339.2 L× 12.4 mol / 122.8 L
n₂ = 4206.08 L.mol /122.8 L
n₂ = 34.3mol
In scientific notation:
3.43×10¹ mol
Answer:
D
Explanation:
It would be D because you are observing the reaction and don’t change anything
0.250 mol/L
<em>Step 1</em>. Write the chemical equation
H2SO4 + 2NaOH → Na2SO4 + 2H2O
<em>Step 2</em>. Calculate the moles of H2SO4
Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)
= 6.25 mmol H2SO4
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)
= 12.5 mmol NaOH
<em>Step 4</em>. Calculate the concentration of the NaOH
[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L
BF3 has a boron atom with three outer-shell electrons in its ground state and three fluorine atoms containing seven outer electrons. Further, if we observe closely, one boron electron is unpaired in the ground state. During the formation of this compound, the 2s orbital and two 2p orbitals hybridize.
Name of the Molecule: Boron Trifluoride
Molecular Formula: BF3
Geometry: Trigonal Planar
