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Answer:12 mol
Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).
6.7mol per 1.3L = 6.7/1.3 mol/L
so in 2.33L = 6.7*2.33/1.3 = 12 mol
Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
