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ra1l [238]
3 years ago
12

1. Enter the complete ionic equation when Li3PO4 and AgNO3 are mixed.2. Enter the net ionic equation when K2SO4 and Na2CO3 are m

ixed.3. Enter the complete ionic equation when K2SO4 and Na2CO3 are mixed.4. Enter the complete ionic equation when Fe(NO3)2 and Na2CO3 are mixed.5. Enter the net ionic equation when Fe(NO3)2 and Na2CO3 are mixed.6. Enter the complete ionic equation when BaCl2 and KOH are mixed.7. Enter the net ionic equation when BaCl2 and KOH are mixed.
Chemistry
1 answer:
cupoosta [38]3 years ago
7 0

Solution :

1. Ionic equation when  $Li_3PO_4$  and $AgNO_3$ are mixed.

  $3 Li^+ (aq) + PO_4^{3-} (aq) + 3Ag^+ (aq) + 3NO_3^-(aq) \rightarrow Ag_3PO_4(s)+3Li^+(aq)+3NO_3^-(aq)$

Net ionic :  $PO_4^{3-}(aq)+3 ag^+(aq) \rightarrow Ag_3PO_4(s) $

2. $K_2SO_4$ and $Na_2CO_3$

  No reaction

3. $K_2SO_4$ and $Na_2CO_3$

  No reaction

4. $Fe(NO_3)_2 \text{ and}\ \ Na_2CO_3$

   $Fe^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+CO_3^{2-}(aq) \rightarrow FeCO_3(s) + 2Na^+(aq) + 2NO_3^-(aq)5. $Fe(NO_3)_2 \text{ and}\ \ Na_2CO_3$

 $Fe^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+CO_3^{2-}(aq) \rightarrow FeCO_3(s) + 2Na^+(aq) + 2NO_3^-(aq)6. $BaCl_2 \text{ and}\ \ KOH$

$Ba^{+2}(aq)+2Cl^-(aq)+2K^+(aq)+2OH^-(aq) \rightarrow Ba(OH)_2(s) + 2K^+(aq) + 2Cl^-(aq)$

7. $BaCl_2 \text{ and}\ \ KOH$

$Ba^{+2}(aq)+2Cl^-(aq)+2K^+(aq)+2OH^-(aq) \rightarrow Ba(OH)_2(s) + 2K^+(aq) + 2Cl^-(aq)$

   

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Assuming steam to be an ideal gas, calculate its specific volume and density at a pressure of 90 lb/in2 and a temperature of 650
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1) Sv = 0.4584 m³/Kg...assuming steam as an ideal gas

% deviation from the values in the steam tables

⇒ % dev = 45 %

2) mass air = 272.617 Kg; assuming air to be an ideal gas

Explanation:

ideal gas:

PV = RTn

molar volume:

⇒ V/n = RT/P

∴ P = 90 psi * ( 0.06895 bar/psi ) = 6.2055 bar

∴ T = 650 F = 343.33 °C = 616.33 K

∴ R = 0.08314 bar.L/mol.K

⇒ V/n = (( 0.08314 )*(616.33 K )) / 6.2055 bar

⇒ V/n = 8.2574 L/mol * ( m³/1000L ) = 8.2574 E-3 m³/mol

specific volume ( Sv ):

∴ Mw = 18.01528 g/mol

⇒ Sv = 8.2574 E-3 m³/mol * ( mol / 18.01528 g ) * ( 1000 g/Kg )

⇒ Sv = 0.4584 m³/Kg

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∴ P = 6.2055 bar ≅ 6 bar → Sv = 0.3157 m³/Kg

⇒ % deviation = (( 0.4584 - 0.3157 ) / 0.3157) * 100

⇒ % dev = 45.2 %; significant value, assuming  steam to be a ideal gas

2) mass air, assuming ideal gas:

∴ V = 20ft * 35ft * 10ft = 7000ft³ * ( 28.3168 L/ft³ ) = 198217.6 L

∴ P = 17 psi * ( 0.06895 bar/psi ) = 1.172 bar

∴ T = 75 °F = 23.89 °C = 296.89 K

∴ R = 0.08314 bar.L/K.mol

⇒ n air = PV/RT = (( 1.172 )*( 198217.6 )) / (( 0.08314 )*( 296.89 ))

⇒ n air = 9411.616 mol air

∴ Mw air = 28.966 g/mol

⇒ mass air = 9411.616 mol * ( 28.966 g/mol ) = 272616.892 g = 272.617 Kg

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