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2 years ago

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For the following aqueous reaction, complete and balance the molecular equation and write a net iconic equatio, making sure to i

nclude the physical states of the compounds:
Potassium carbonate + Strontium nitrate

1 answer:

pav-90 [236]2 years ago

6 0

Answer:

Balance molecular equation:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

Explanation:

Potassium carbonate = K2CO3

Strontium nitrate = Sr(NO3)2

Chemical equation:

K2CO3 + Sr(NO3)2 → SrCO3 + KNO3

Balance chemical equation with physical states:

K2CO3(aq) + Sr(NO3)2(aq) → SrCO3(s) + 2KNO3(aq)

Ionic equation:

2K+(aq) + CO3∧-2(aq) + Sr∧+2(aq) + 2NO∧-3(aq) → SrCO3(s) + 2K+(aq) + 2NO∧-3(aq)

Net ionic equation:

CO3∧-2(aq) + Sr∧+2(aq) → SrCO3(s)

2K+ and 2NO∧-3 ions are spectator ions that's way these are not written in net ionic equation.

Spectator ions:

These are the ions that are present same on both side of chemical reaction and does not effect the equilibrium.

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3 years 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

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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

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

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

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The generalized Pitzer correlation is :

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

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

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

$T__{\gamma}} = 0.808$

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

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

$P__{\gamma}} = 0.0816$

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

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

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

$B_1 = -0.282$

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

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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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