Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.

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:

Equivalentes KOH:

0.100L * (1eq / L) = 0.100eq = 0.100moles

Gramos KOH -Masa molar: 56.1056g/mol-:

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

KOH 90%:

5.61g KOH * (100g KOH 90% / 90g KOH) =

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explanation:

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.