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mihalych1998 [28]
3 years ago
15

Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by rea

cting with CO2 in the air to produce sodium carbonate (Na2CO3) and O2. 2 Na2O2(s) + 2 CO2(g) → 2 Na2CO3(s) + O2(g) What volume (in liters) of CO2 can be consumed at STP by 435 g Na2O2?
Chemistry
1 answer:
bazaltina [42]3 years ago
6 0

 The volume  (in liters)  of CO₂  that can  be consumed at STP by 435 g Na₂O₂  is  125 L of Co₂

<u><em>calculation</em></u>

2Na₂O₂(s)  +2 CO₂ (g)→  2 Na₂CO₃(s)  + O₂(g)

Step 1 : find the moles  of Na₂O₂

moles = mass÷  molar mass

from periodic table the  molar mass  of Na₂O₂ = (23 x2) +( 16 x2) = 78 g/mol

 moles= 435 g÷ 78 g/mol = 5.58 moles

Step 2: use the mole ratio to determine the moles of CO₂

from given equation  Na₂O₂ : CO₂ =2 :2 =1:1

Therefore the  moles of CO₂  is also = 5.58 moles

Step 3: find the volume  of CO₂  at STP

that is at STP      1  mole of a gas = 22.4 L

                          5.58 moles = ? l

<em>by cross multiplication</em>

= (5.58 moles x 22.4 L) / 1 mole = 125 L

 

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Average atomic mass listed for nitrogen in the periodic table is 14

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3 0
3 years ago
Calcium oxide and oxygen gas are produced by the thermal decomposition of limestone in the reaction CaCO (s) CaO(s) + CO (g). Wh
kozerog [31]
The mass of lime that  can  be produced  from  4.510 Kg of limestone  is calculated  as  below

calculate the moles  of CaCO3  used

that is  moles =mass/molar mass
convert  Kg  to g  = 4.510 x1000 =4510g
=  4510 / 100 =45.10 moles
CaCO3 = CaO  +O2

by use of mole ratio between CaCO3  to CaO  (1:1) the  moles of CaO  is also= 45.10 moles
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6 0
3 years ago
Read 2 more answers
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
Is this molecule polar or nonpolar?
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3 0
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Answer:

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