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:
2B2 + 3O2 → 2B2O3
Explanation:
Balance The Equation: B2 + O2 = B2O3
1. Label Each Compound With a Variable
aB2 + bO2 = cB2O3
2. Create a System of Equations, One Per Element
B: 2a + 0b = 2c
O: 0a + 2b = 3c
3. Solve For All Variables (using substitution, gauss elimination, or a calculator)
a = 2
b = 3
c = 2
4. Substitute Coefficients and Verify Result
2B2 + 3O2 = 2B2O3
L R
B: 4 4 ✔️
O: 6 6 ✔️
hope this helps!
9.96*10⁻²² is the answer, First you find the molar mass of gold, then you turn atoms to moles then moles to grams. There you go you're done!<span />
Heterogeneous
The mixture has more than one phase of matter in it.
Answer:
13896g
Explanation:
volume = 4.50×8.00×20.00 = 720 cm³
mass = density × volume
mass= 19.3 × 720 = 13896g
