please solve ans show work Constants for mercury at 1 atm heat capacity of Hg( 2

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please solve ans show work

Constants for mercury at 1 atm heat capacity of Hg( 28.0 J/(mol K) melting point enthalpy of fusion Calculate the heat energy released when 21.9 g of liquid mercury at 25.00°C is convertedto solid mercury at its melting point. 234.32 K 2.29 kJ/mol Number k.J Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas whern heated, thus inflating the bag 2NaN3(s) 2Na(s) +3N2(g) Calculate the value of work, w, for the following system if 25.7 g of NaN3 reacts completely at 1.00 atm and 22 °0C Number For each of the reactions, identify other quantities that are equal to enthalpy of reaction, rrxn. 2.. CH,(g) C(g) +4H(g) enthalpy of atom combination for CH4 enthalpy of atomization of CH4 enthalpy of combustion of CH4 enthalpy of formation of C(g) 4xbond energy of C-H enthalpy of atom combination for CH4 enthalpy of atomization of CH4 enthalpy of combustion of C enthalpy of formation of CH4(g) 4x bond energy of C-H Calculate the standard enthalpy change for the following reaction at 25 °C. Standard enthalpy of formation values can be found in this tabl HCI(g) NaOH(s) NaCl(s):H10( Number k.J

Explanation / Answer

SOLUTION:

1.

mercury moles = mass/ Molar mass ofit = 21.9 /200.59 = 0.1092

Heat released in cooling mercury from 25 to its Melting point i.e -38.829 C

     = molar heat capacity of mercury x moles of mercury x temperature change

    = 27.983 J/molK x ( 0.1092 moles) x ( 25-(-38.829)

      = 195 J = 0.195 KJ

Heat released = molar heat of fusion x moles of mercury

     = 2.29 KJ/mol x 0.1092 moles ( where 2.29 KJ/mol is molar heat of fusion of mercury taken from web)

       = 0.25 KJ

now total heat released for the process = sum of above two process

         = 0.195+0.25 = 0.445 KJ

thus 0.445 KJ of heat is released when mercury at 25 is cooled and converted to solid at its melting point

2.

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2NaN3(s)-->2Na(s)+3N2(g)

Calculate the value of work, w, for the following system if 25.7 g of NaN3 reacts completely at 1.00 atm and 22 °C.

mol of NaN3 = mass/MW = 25.7/65.0099 = 0.395324 mol of NaN3

now, relate mol of NaN3 to mol of gases, N2

2 mol of NAN3 = 3 mol of N2

0.395324mol --> 3/2* 0.395324 = 0.592986 mol of N2

assume this is the only gas

The Work equation

W = -P*dV

W = -P*(Vf-Vi)

Assume:

VF >>> Vi, since gas > solids

so

Vf = Vgas

W = -P*(Vf)

Vf--> apply ideal gas law

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Tmperature

R = ideal gas constant

V = nRT/P

T = 273 + 23 = 296 K

V = (0.592986 )(0.082)(296)/(1)

V = 14.3929 L = 14.3929*10^-3 m3

W = -P*(Vf)

P = 1 atm = 101325 Pa

W = -101325*(14.3929*10^-3) m3Pa

W = -1458.3605 J

W = -1.45 kJ/mol

note:as per chegg policy we will give first full question only but i am giving second one also hope this helps you thank you

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