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drek231 [11]
2 years ago
5

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

Chemistry
1 answer:
attashe74 [19]2 years ago
8 0

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>
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The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent
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Answer:

<h2>0.06 % </h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

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<h3>0.06 % </h3>

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4 0
3 years ago
1) If the pressure on 2.6 liters of a gas at 860 Torr is increased to 1000 Torr,
LenKa [72]

2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.

Explanation:

Data given:

Initial volume of the gas V1 = 2.6 liters

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final pressure on the gas P2 = 1000 Torr 1.315 atm

final volume of the gas after pressure change V2 =?

From the data given above, the law used is :

Boyles Law equation:

P1V1 = P2V2

V2 = P1V1/P2

   = 1.13 X 2.6/ 1.31

  = 2.24 Liters

If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.

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Why is the r.a.m. Value used?
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Answer:

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