C5H12 + 8 O2 → 5 CO2 + 6 H2O
8 molecules of O2 are required.
Answer:
737.52 mL de agua
Explanation:
En este caso solo debes usar la expresión de molaridad de una solución la cual es:
M = moles / V
Donde:
V: Volumen de solución.
Como queremos saber la cantidad de agua, queremos saber en otras palabras cual es la cantidad de solvente que se utilizó para preparar los 800 mL de disolución.
Una disolución se prepara con un soluto y solvente. El soluto lo tenemos, que es el nitrato de plata. Con la expresión de arriba, calculamos los moles de soluto, y luego su masa. Posteriormente, calculamos el volumen con la densidad, y finalmente podremos calcular el solvente de esta forma:
V ste = Vsol - Vsto
Primero calcularemos los moles de soluto:
moles = M * V
moles = 2 * 0.800 = 1.6 moles
Con estos moles, se calcula la masa usando el peso molecular reportado que es 169.87 g/mol:
m = moles * PM
m = 1.6 * 169.87 = 271.792 g
Ahora usando el valor de la densidad, calcularemos el volumen de soluto empleado:
d = m/V
V = m/d
V = 271.792 / 4.35
V = 62.48 mL
Finalmente, la cantidad de agua necesaria es:
V agua = 800 - 62.48
V agua = 737.52 mL
Answer:
Explanation:
For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..
Answer: The value of 
for the reaction is, -2512.4 kJ
Explanation:
The chemical equation for the combustion of acetylene follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(4\times (-393.5))+(2\times (-241.8))]-[(2\times (227.4)+(5\times (0))]\\\\\Delta H^o_{rxn}=-2512.4kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%284%5Ctimes%20%28-393.5%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28227.4%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-2512.4kJ)
Therefore, the value of for the reaction is, -2512.4 kJ
<span>One that has 30.0 grams of NaCl dissolved in 100 grams of water is more concentrated than one that has 15.0 grams of NaCl dissolved in 100 grams of water. A is the correct answer.</span>
