The ion N³⁻ is called the azide ion. In its neutral state, it occurs as the element Nitrogen. The atomic number of Nitrogen is 7. When it turns into an anion (negatively charged ion), it gains 3 more electrons. That's why its net charge becomes -3. It means that the protons is still 7, but the electrons are now 10.
Overall charge = +7 + -10 = -3
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>
An Olympic decoration is granted to effective contenders at one of the Olympic Amusements. There are three classes of decoration: gold, granted to the victor; silver, granted to the first sprinter up; and bronze, granted to the second sprinter up.
Is there any answer choices before i get started on working out the problem
