Please solve the followiing ASAP Calculate the volume (in liters) occupied by 1.
ID: 839868 • Letter: P
Question
Please solve the followiing ASAP
Calculate the volume (in liters) occupied by 1.03 moles of nitric oxide (NO) at 7.32 atm and 67 degree C. The equation for the metabolic breakdown of glucose (C6H1206) is the same as the equation for the combustion of glucose in air Calculate the volume of C02 produced at 37 degree C and 1.00 atm when 3.58 g of glucose is used up in the reaction. A gas occupying a volume of 691.0 mL at a pressure of 0.867 atm is allowed to expand at constant temperature until its pressure reaches 0.550 atm. What is its final volume?Explanation / Answer
1) apply
PV= nRT
V = nRT / P
V= 1.03 x 0.0821 x 340 / 7.32
V = 3.927 L
so the volume is 3.927 L
2)
First caculate at standard conditions
form the given reaction
moles of C02 produced = 6 x moles of glucose
moles of glucose = mass /molar mass
moles of glucose = 3.58 /180
moles of glucose = 0.01988
moles of C02 produced = 6 x 0.01988
moles of C02 = 0.11933
now apply
V = nRT/P
V = 0.11933 x 0.0821 x 273 / 1
V =2.674
now at T = 37 C and P= 1 atm
pressure is constant
So
V1/T1 = V2 /T2
V2 = v1 T2 / T1
V2 = 2.674 x 310 / 273
V2 = 3.037
so the volume of C02 is 3.037 L
3) Given constant temp
apply boyles law
P1V1 =P2 V2
V2 = P1V1 / P2
V2 = 0.867 x 691 / 0.55
V2 = 1089.27 ml
so the final volume is 1089.27 ml
4) Given constant pressure
apply charles law
V1/T1 = v2 /T2
V2 = V1 T2 /T1
V2 = 32.9 x 339 / 301
V2 = 37.05
so the final volume is 37.05 L
5)
apply n = PV /RT
n = 4.18 x 2.12 / 0.0821 x 300
n = 0.36
so the moles are 0.36
6) Apply V = nRT/ P
V = 5.97 x 0.0821 x 404 / 9.57
V = 20.69
so the volume is 20.69 L
7) at STP
1 mole of gas occupies 22.4 L
moles of gas = volume / 22.4
moles of gas = 0.219 / 22.4
moles of gas = 9.776 x 10-3
molar mass = mass /moles
molar mass = 0.55 / 9.776 x 10-3
molar mass = 56.25
so the molar mass of gas is 56.25 g/mol
8)
a) density d = mass /volume
d = 4.45 / 2.05
d= 2.17 g /L
b) PV = nRT
PV = mRT/ M
PM = ( m/V) RT
PM = dRT
M = dRT/P
M = 2.17 x 0.0821 x 300 / 1
M = 53.46
so the molar mass is 53.46 g /mol
9)
2NaHC03 -----> Na2C03 + C02 + H20
so moles of C02 = moles of NaHC03 /2
moles of NaHC03 = 5.92 /84
moles of NaHC03 = 0.07
moles of C02 = 0.07/2
moles of C02 = 0.035
V = nRT/P
V= 0.035 x 0.0821 x 273 / 1
V = 0.7898
but this is at standard conditions
now apply
P1 V1 / T1 = p2 V2 /T2
V2 = P1 V1 t2 / P2 T1
V2 = 1 x 0.7898 x 453 / 1.34 x 273
V2 = 0.978
so the volume of C02 is 0.978 L
10)
heat lost by hot body = heat gained by cold body
heat lost by steel = heat gained by water
Ms Ss dTs = Mw Sw dTw
30.45 x 0.474 ( 104.89 - T) = 116.9 x 4.184 x ( T - 19.55 )
T = 21.99
so the final temperature is 21.99 C
11) dHrxn = dHf products - dHf reactants
dH rxn = 6 x dH C02 + 6 x dHf H20 - dHf C6H12 - 9 x dHF 02
dH rxn = - 6 x 393.5 - 6 x 285.8 + 151.9 -0
dH rxn = -3923.9 kJ
so
dH rxn = -3923.9 kJ/mol
12) The final eqaution = second equation + 2 x third equation - first equation
so
dH rxn final equation = dHrxn fo second equation + 2 x dHrxn third equation - dhrxn first equation
dH rxn final = -393.5 - 2 x 285.8 + 726.4
dH rxn final = -238.7
so dHf CH3OH = -238.7 kJ /mol
13) moles of P4 = mass /molar mass
moles of P4 = 411 / 124
moles = 3.314
from the reaction
Heat evolved = dH /moles of P4
Heat evolved = -3013 / 3.314
Heat evolved = -909.03
Heat evolved = -909.03 kJ
14)
the required equation is
C + S2 ---> CS2
final equation = reaction 1 + 2 x reaction2 - reaction 3
dHf CS2 = -393.5 - 2 x 296.4 + 1073.6
dHf CS2 = 87.3 kJ /mol
so the answer is 87.3 kJ /mol
15) from the reaction
2 moles of ZnS gives -879 kJ
Heat per mole of ZnS = Heat /moles
Heat per mole of ZnS = -879/2
Heat per mole of ZnS = -439.5
Heat per gram of ZnS = Heat per mole / molar mass
Heat per gram = -439.5 / 97.474
Heat per gram = -4.509
heat per gram is -4.509 kJ
16) Wave length = velocity / frequancy
wave length = 3 x 108 / 4.42 x 106
wave length = 67.87 m
so the wave length is 67.87 m
17) n = 3
L = 1
18)
values of m range from -L to + L
for 3p orbital L=1
so m values are from -1 to +1
m values are -1 ,0, +1
19) Electron 1 : n=1 : L=0 : M =0
Electron 2 : n=1 : L=0 : M =0 : Ms = -1/2
Electron 3 : n=2 : L=0 : M =0 : Ms = +1/2
Electron 4 : n=2 : L=0 : M =0
Electron 5 : n=2 : L=1 : Ml = -1,0,+1 : Ms = + 1/2
20 ) frequency = 3 x 108 / 541 x 10-9
frequency = 5.54 x 10^14 Hz
wave length = 3 x 108 / 4.29 x 109
wave length = 6.99 x 10^7 nm
21) Given L= 1
so m values are from -1 to 1
m values are -1,0,1
b) for n we have
L = 0 to n-1
L = 0,1,2