17. What is the coefficient of H0 when the following equation is smallest set of

Question

17. What is the coefficient of H0 when the following equation is smallest set of whole numbers A) 3 B) 4 C)6 D) 12 E) 24 ) many molecules of N gas can be present in a 2.5 L. flask at 50 C and 650 mmHg? A) 2.1 x 10-3 molecules 18. How D) E) 3.6×1025 molecules 0.081 molecules B) 4.9 x 1022 molecules C) 3.1 x 10 molecules 19. Identify the major ionic species present in an aqueous solution of H:SO4. O3 (plus H neutral species) B H, Or. C) 2H'. So, 40" D) H, HSO E) 2H, SO 20. Predict the products of the following single replacement reaction Fe(s) + CuSO4(aq) A) Cu(s) +FeSO4(ag) B) Fe(s) +Cu(s) +SO4(aq) C) CuS(s)+ Fe SO(aq) D) FeCuSO (aq) E) FeO(s)+CuSOs(ag) 21. Calculate the mass, in grams, of 2.74 L of CO gas measured at 33°C and 945 mmiig. A) 0.263g B) 2.46g C) 3.80g D) 35.2g E) 206g 22. Ferrocene, Fe(C,Hs)-(s), can be prepared by reacting 3.0 g of FeCl2(s) with an equal mass of cyclopentadiene, CHol), and an excess of KOH, as shown in the following reaction The theoretical yield of Fe(CsHsh was 4.2g. If the actual yield of ferrocene is 2.7g, calculate the percent yield. A) 6496 B) 0.64% C) 6.4% D) 1.55% E) 155.6% 23. A 1.2 L flask contains 0.500 mole of ammonia (NH) at I50°C. Calculate the pressure of the ammonia inside the flask. A) 691 x 10 atm B) 5.13 atm C) 12.2 atm D) 14.5 atm E) 22.4 atm

Explanation / Answer

17) Al4C3 + 12H2O 4Al(OH)3 + CH4

The answer will be D) 12.

18) P = 650 mm Hg =( 650 /760) atm = 0.855 atm

T = (50 + 273) K = 323 K

Now, PV= nRT

=> n = PV/RT = 0.855 atm x 2.5 L / 0.08206 Latm /mol K x 323 K

=> n = (2.1375/26.50538)moles = 0.0806 moles = 0.081 moles

Now , 0.081 mol x (6.022 x 1023 molecules / mol) = 4.89 x 1022 molecules = 4.9 x 1022 molecules

The answer will be B) 4.9 x 1022 molecules

19) H2SO4(aq) ---> 2H+(aq) + SO42-(aq)

The major ionic species will be , 2H+ and SO42-. The answer will be E) 2H+ , SO42-

20) Fe(s) + CuSO4 (aq) ----> FeSO4 (aq) + Cu(s)

The answer will be A) Cu(s) + FeSO4(aq)

21)  P= 945/760=1.24 atm

T = 33 + 273 =306 K

moles(n) CO = PV/RT = 1.24 atm x 2.74 L / 0.08206 Latm/mol K x 306 K=(3.3976 / 25.11036) = 0.135 moles

Mass CO = 0.135 mol x 28.01 g/mol=3.789 g = 3.80 g

The answer will be C) 3.80 g

23) PV = nRT

P = nRT / V

T=1500C = (150 + 273) K = 423 K

P = 0.500 mol x 0.08206 Latm/ mol K x 423 K / 1.2 L

=>P = (17.35569/ 1.2) atm = 14.46 atm = 14.5 atm

The answer will be D) 14.5 atm

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