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1 year ago

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Zeolite A, [(AlO₂)₁₂(SiO₂)₁₂] .27 H₂O, is used to soften water by replacing Ca²⁺ and Mg²⁺ with Na⁺. Hard water from a certain so

urce is 4.5 × 10⁻³MCa²⁺ and 9.2 × 10⁻⁴MMg²⁺, and a pipe delivers 25,000 L of this hard water per day. What mass (in kg) of zeolite A is needed to soften a week’s supply of the water? (Assume zeolite A loses its capacity to exchange ions when 85 mol % of its Na1 has been lost.)

1 answer:

statuscvo [17]1 year ago

7 0

47.43 kg of Zeolite A is needed to soften a week's supply of the water.

<h3>What is Zeolites ?</h3>

Zeolites are 3-dimensional crystalline solids which are formed either synthesized or can be naturally. Zeolites belongs to class microporous solids.

Given

Hard water supply per day = 25,000L

Ca⁺² concentration = 0.0045 mol/L

Total mole of Ca⁺² = 25,000 × 0.0045

= 112.5 mol

Mass of Ca⁺² in hard water = Mole × Molar mass

= 112.5 × 40.08 g

= 4509 g

Concentration of Mg⁺² = 0.00092 mol/L

Total mole of Mg⁺² = 0.00092 × 25000

= 23

Mass of Mg⁺² in Hard water = Mole × Molar mass

= 23 g × 18.04

= 414.92 g

Molar mass of Zeolite = 2190g/mol

No of mole of Ca⁺² removed by 1 mole of Zeolite A = 6

Mass of Ca⁺² removed by 2190 g of Zeolite A = 240.48g

Mass of Zeolite A required to remove 4509 g of Ca⁺² = 33.73kg

Assume efficiency of Zeolite is 85%

To remove Ca⁺² the mass of Zeolite A = 39.68kg

No of mole of Mg⁺² removed by 1 mole of Zeolite A = 6

Mass of Mg⁺² removed by 2190g of Zeolite A = 145.86g

To remove 414.92 g of Mg⁺² the mass of Zeolite = 6.59kg

Assume the 85% as the efficient of Zeolite A

Mass of Zeolite A required to remove Mg⁺² = 7.75kg

Total mass of Zeolite A = 39.68kg + 7.75kg

= 47.43kg

Thus from the above conclusion we can say that 47.43 kg of Zeolite A is needed to soften a week's supply of the water.

Learn more about the Zeolites here: brainly.com/question/14976531

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alexandr402 [8]

Answer: The process is diffusion.

Explanation:

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1 year ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

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When 1 mol of KBr(s) decomposes to its elements, 394 kJ of heat is absorbed. (a) Write a balanced thermochemical equation.

VladimirAG [237]

The balanced thermochemical equation is

KBr ------- K + 1/2 Br2

<h3>What is thermochemical equation? </h3>

A Thermochemical Equation is defined as the balanced stoichiometric chemical equation which includes the enthalpy change, ΔH.

The chemical equation for the decomposition of potassium bromide to its constituent elements bromine ans potassium :

KBr ----- K + Br2

The balanced thermochemical equation of the decomposition of potassium bromide to its constituent elements potassium and bromide as follows

KBr ------- K + 1/2 Br2

As the heat is absorbed in this reaction therefore, heat is positive.

Thus, we concluded that the balanced thermochemical equation is

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brainly.com/question/2733624

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