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AfilCa
15 days ago
14

Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^

2+
Chemistry
1 answer:
lorasvet [2.5K]15 days ago
5 0
Fe 3+ + SCN- --> FeSCN 2+ 

<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>

<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>
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Answer: The right choice is (c) application of both a mobile phase and a stationary phase.

Explanation:

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Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.

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A compound composed of only carbon and chlorine is 85.5% chlorine by mass. propose a lewis structure for the lightest of the pos
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The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

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n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

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n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

4 0
1 month ago
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The temperature of the gas rises.

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11 days ago
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7 0
1 month ago
(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo
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Answer:

La masa molar del compuesto es: 168.82 g/mol

La masa molar del gas es: 16.38 g/mol

Explanation:

(a)

Utilizando la ecuación de gases ideales:

PV=nRT

donde,

P es la presión

V es el volumen

n es el número de moles

T es la temperatura

R es la constante de los gases, cuyo valor es = 0.0821 L.atm/K.mol

Además,

Moles = masa (m) / Masa molar (M)

La densidad (d) = Masa (m) / Volumen (V)

Así, la ecuación de gases ideales se puede expresar como:

PM=dRt

Dado que:-

Presión = 20 kPa = 20000 Pa

La expresión para la conversión de presión en Pascal a presión en atm se muestra a continuación:

P (Pa) = \frac {1}{101325} P (atm)

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Masa molar =?

Aplica la fórmula:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

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(b)

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Presión = 152 Torr

Temperatura = 298 K

Volumen = 250 cm³ = 0.25 L

Utilizando la ecuación de gases ideales:

PV=nRT

R = 62.3637\text{torr}mol^{-1}K^{-1}

Aplicando la fórmula:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

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Dado que:

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Masa molar =?

La fórmula para calcular los moles se muestra a continuación:

moles = \frac{Mass\ taken}{Molar\ mass}

Así,

0.002045\ moles&#10;= \frac{0.0335\ g}{Molar\ mass}

La masa molar del gas es: 16.38 g/mol

5 0
1 month ago
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