26. Adding 5.44 g of NHaNOxo to 150.0 g of water in a coffee-cup calorimeter (wi

Question

26. Adding 5.44 g of NHaNOxo to 150.0 g of water in a coffee-cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6°C to the final temperature of 16.2°C. Calculate the enthalpy change for dissolving NH.NO3o in water, in ki/mol. Assume the solution (whose mass is 155.4 g) has a specific heat capacity of 4.20 i/ (g x°C) NH4NOsolie) NH NO,laqueous) Find AH° in kJ/mol. 4 pts. Enthalpy change per mol for dissoving NHaNO; ki/mol 27. You add 100.0-g of water at 60.0°C to 100.0-g of ice at 0.00°C. Some of the ice melts and cools the water to 0.00°C. When the ice and water mixture reaches thermal equilibrium at 0.00°C, how much ice has melted? ISpecific heat capacity of water 4.184j/(g x °C) and heat of fusion of water 333J/ 4 pts. Amount of ice that had melted - 28. What is the total pressure in atmospheres of a gas mixture that contains 1.00 g of hydrogen (H2), and 8.00 g of argon (Ar) in a 3.00 L container at 27°C? 8 pts. a) Total pressures at b) What are the partial pressure of each the 2 gases? Partial pressure of H2 Partial pressure of Ar- Fu I Ph cal constants and important equations, Avogadro's number 6.02214x10" units /mole Units can be molecules, atoms, formula weights, or other entities. Universal gas constant R 8.2058x102/L atml/imol Kl- 0.082058 (L Molarity = M: moles of solute/liter of solution and M-v = moles atm)/imol K Dilution of solutions: M. . V-M2·V2 (You must know 3 of the 4 components) Density Mass/volume (Mass of compound) x (1 mole/formula weight)- moles of compound (Moles of compound) x (6.022 14x1023 units / mole) . # molecules or " atoms or # formula units T (in K) = T (in °C) + 273.15 1 atm 101.325 kilopascals [kPa]s 101,325 Pa-760 mm Hg 760 Torr 1.01325 Barr man" molar mass x P mass Pv-nRT PV nRT Molar mass Standard molar volume at STP (and 1 mol) 22.4144 P. ((moles X)/ (total moles)) x (total pressure) n= n a D- Molar mass RT P.VI/T -P:V:/Tz ??,ystem-(-) ?? surroundings ??: standard heat at STP (e) of formation () of a compound from its elerments in their standard states. Some element standard states are: Paist, S Ozie Hzgh, F2e N2tel, etc.

Explanation / Answer

Solution:- (26) Q = m c delta T

delta T = 18.6 - 16.2 = 2.4 degree C

m = 155.4 g

So, Q = 155.4 g x 4.20 J/(g. degree C) x 2.4 degree C

Q = 1566.432 J or 1.57 kJ

delta H = Q/n

where n is the number of moles of NH4NO3.

5.44 g x (1mol/80.04 g) = 0.0680 mol NH4NO3

delta H = 1.57 kJ/0.0680 mol

delta H = 23.1 kJ/mol

(27):- mass of water = 100.0g

initial temperature of water = 60.0 degree C

Final temperature = 0.00 degree C

change in temperature for water, delta T = -60.0 degree C

Heat lost by water, Qwater = 100.0 g x 4.184 J/(g. degree C) x (-60.0 degree C)

Qwater = -25104 J

heat given = - heat taken

let's say X grams of ice melted.

Qice = m*delta Hfus

where delta Hfus is the enthalpy of fusion and it's value is 333 J/g.

Since, Qwater = - Qice

So, 25104 J = X(333 J/g)

X = 25104 J x (1g/333J)

X = 75.4 g

So, out 100.0 g,only 75.4 g of ice were melted.

(28) - (a) 1.00 g H2 x (1mol/2.02g) = 0.495 mol H2

8.00 g Ar x (1mol/39.948 g) = 0.200 mol Ar

Total moles of gas = 0.495 + 0.200 = 0.695

using PV = nRT, we can calculate the total pressure.

P = nRT/V

P = (0.695mol) x (0.0821atm.L/mol.K)x (300K)/3.00L

P = 5.71 atm

(b) mole fraction of H2 = 0.495/0.695 = 0.712

partial pressure of H2 = mole fraction x total pressure

partial pressure of H2 = 0.712 x 5.71 atm = 4.07 atm

partial pressure of Ar = 5.71 atm - 4.07 atm = 1.64 atm

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