<em>V= 110mL = 110cm³ = 0,11dm³</em>
<em>C = 1,244 mol/L = 1,244 mol/dm³</em>
C = n/V
n = 1,244×0,11
<u>n = 0,13684 moles</u>
<em>mCa(OH)₂ = 74 g/mol</em>
1 mole Ca(OH)₂ ------------ 74g
0,13684 ---------------------- X
X = 74×0,13684
<u>X = 10,12616g</u>
:)
Answer: the boiling point elevation constant is 
Explanation:
Elevation in boiling point is given by:
= Elevation in boling point
i= vant hoff factor = 1 (for non electrolyte)
=boiling point constant = ?
m= molality
Weight of solvent (diethylether)= 330 g = 0.33 kg
Molar mass of solute (benzophenone)= 182 g/mol
Mass of solute (benzophenone) = 38.2 g
Thus the boiling point elevation constant is
Answer:
3.824 atm
Explanation:
From the ideal gas equation
P = mRT/MW × V
m is mass of testosterone = 12.9 g
R is gas constant = 82.057 cm^3.atm/mol.K
T is temperature of benzene solution = 298 K
MW is molecular weight of testosterone = 288.40 g/mol
V is volume of benzene solution = 286 ml = 286 cm^3
P = 12.9×82.057×298/288.4×286 = 3.824 atm
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>
