2-what is the mass of carbon needed to react with 4,0 kg of Solid titanum onde (

Question

2-what is the mass of carbon needed to react with 4,0 kg of Solid titanum onde (TO)#25% of the carbon must be recovered in the reaction of production of solid titanium (T) and gaseous carbon monoxide (CO 2-Fil in the blanks Compound Nal CHon .15 HO.15 molality percent mole fraction 0.15 5.0 100 CH,COOH 00183 3- Ammonium nitrate (NH NO-4s) ) decomposes into nitrogen, oxygen and water. 12 g of ammonium nitrate decompose entirely at 300 "C with AH-236. How much heat is liberated? How much work would be done by the gas? - What volume of gas would be produced at a pressure of 1 bar? 4 Calculate the standard molar entropy change for the formation of HCNE) from the elements at 25c 5-Aneobrium miture in 0S L flask contains 070 mol HKg) and 0.10 mol each of 11g) and HIE·lf030 mol Ha 0.60 moll h and 0.20 mol Hil are added to this equilibrium mixture how many moles of each gas will be present when equillibrium is reestablished? 6-What is the change of pH upon addition of 20.0 ml of HCI 3.0 M to 750.ml of the following -pure water -0.1 M formic acid 0.1M acetate of sodium - a buffer made of 0.1M potassium nitrite and 0.08 M nitrous acid. 7- HON dissociates into CN and protons a) What is the equilibrium constant at T-298.15 K? b) At what pH will the reation move to the left if the solution contains 0.1 M HCN? 8 50 ml of a monoprotic weak acid are titrated with NaOH 0.1 M Equivalence is reached at 27.0 mi delivered. The pi measured at equivalence is 6.15 a) What is the initial amount of the weak acid? b) What is the concentration of the weak acid? c) What is the weak acid? d) Calculate the initial pH e) Calculate the pH after 5.0 of titrant have been delivered calculate the pH after 30.0 ml of titrant have been delivered What is the emf of a cell consisting of a half cell a PuHTM half cell if fPb_] = 0.10 M, pr-0.050 M and Pic-1.5bar?

Explanation / Answer

2. NaI

molality = 0.15 m

molality = mass of solute/kg of solvent

solvent is water (density = 1 g/cm^3)

g/kg of NaI = 0.15 x 149.89 = 22.4835 g/kg

moles of NaI = 0.15 mols

So, mass of solvent = 1000 g

moles of water = 1000/18.015 = 55.51 mols

moles fraction of NaI = 0.15/0.15 + 55.51 = 2.7 x 10^-3

mass fraction of NaI = 22.4835/22.4835 + 1000 = 0.022 mols

C2H5OH

mass fraction = 5

mass of C2H5OH = 5 g

mass of water = 100 - 5 = 95 g

moles of C2H5OH = 5/46.07 = 0.1085 mols

moles of water = 95/18.015 = 5.2734 mols

moles fraction of C2H5OH = 0.02

molality of solution = 0.1085/0.095 = 1.142 m

C12H22O11

you may do the same as we did for NaI

Same can be done for CH3COOH

KNO3

mole fraction = 10

means we have 10 moles of

moles of solvent = 100 - 10 = 90 mols

mass of KNO3 = 10 x 101.1032 = 1011.032 g

mass of solvent = 90 x 18.015 = 1621.35 g

mass fraction KNO3 = 1011.032/1011.032 + 1621.35 = 0.38

molality = 10/1.62135 = 6.17

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