Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
-81.5 degrees C or 191.5 K
Explanation:
We want to use Charles' gas law: V/T = V/T
Our initial volume is 3.20 L, and our initial temperature is 125 degrees C, or 125 + 273 = 398 degrees Kelvin.
Our new Volume is 1.54 L, but we don't know what the temperature is. So, we use the equation:
3.20 L / 398 K = 1.54 L / T ⇒ Solving for T, we get: T = 191.5 K
If we want this in degrees Celsius, we subtract 273: 191.5 - 273 = -81.5 degrees C
Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30
$724.73 this would be the answer because if you subtract 320.50 and 86.10 from the 1056.33 then add 75 you get 724.73
