To determine the amount of a substance in units of moles from units of grams, we need to determine the molar mass of the substance. <span>The </span>molar mass<span> is the </span>mass<span> of a given chemical element or chemical compound (g) divided by the amount of substance (mol). For CuF2, the molar mass </span><span>101.543 g/mol. We calculate as follows:
100.0 g CuF2 ( 1 mol / 101.543 g) = 0.98 mol CuF2</span>
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>
<span>Synthesis, decomposition, single replacement and double replacement.</span>
Answer:
Carbonation. When you think of carbonation, think carbon
Oxidation. Oxygen causes oxidation.
Hydration. This isn't the hydration used in your body, but it's similar.
Explanation:
Answer:

Explanation:
Hello there!
In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

Best regards!