Answer: 19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2
solution:
Hydration is the addition of water; hydrogenation is the addition of hydrogen.
desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??
knowns:
__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole
__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole
___________2(1/2 O2 + H2 -------------> H2O)___________2*dHox = 2(-285.8 kJ/mole)
Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem
thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.
Answer:
premium: 91 octane rating
Explanation:
Octane number refers to the percentage or volume fraction of isooctane in a fuel.
The octane number gives a picture of how safe a fuel is for an engine. The higher the octane rating the lesser the tendency of the fuel to cause knocking of the engine.
The type of gasoline with the highest percentage of octane among the options is premium.
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
