Answer:
17.6 grams of nitrogen gas
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>
Answer:
it holds reactant molecules in a good position for then to react
The reaction of an Arrhenius acid with an Arrhenius base produces water and <span>A) a salt</span>
The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms
<h3>Avogadro's hypothesis </h3>
1 mole of Mg = 6.02×10²³ atoms
<h3>How to determine the atoms in 0.58 mole of Mg </h3>
1 mole of Mg = 6.02×10²³ atoms
Therefore,
0.58 mole of Mg = 0.58 × 6.02×10²³
0.58 mole of Mg = 3.49×10²³ atoms
Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg
Learn more about Avogadro's number:
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