Free Home Work Essay Sample

Qn 1.2. this is chemical change because atoms of an element have been broken down and used to join up the other compound. Breaking down of atoms and the joining with other atoms involve the consumption of energy and change in chemical state which is irreversible by chemical means. This makes it a chemical change.

Qn 1.5. a) Measurement (ii) is both accurate and precise because the targets are within the bull's-eye area, the red spot. This shows that the person was accurate by hitting the right target and hitting other points close to the right target. Accuracy and precision is observed.

b). Measurement (i) is precise but inaccurate. It is inaccurate because the hits are away from the right target. However, they are precise because the hits are within the first target thereby making it more precise.

c). Measurement (iii) are imprecise but can yield an accurate average. The imprecision is brought about by hit targets are far away from each other. However, accuracy arises because the targets are at somewhat equidistant measurement from the centre. An average would therefore result as if the centre target has been hit.

Qn. 1.12. a) Air is a mixture because it is made up of various gasses like nitrogen, oxygen, carbon dioxide among others. It is however heterogeneous because the ratio of composition of the elements involved are not equivalent. 

b). tomato juice is a mixture because it comprises of water and the red pigmentation arising from the juice. This also makes the substance to be heterogeneous.

c) Iodine crystals are a pure substance because they arise from the only single element, iodine. This also makes it to be homogenous because the only visible components are the crystals.

d). sand is a pure substance but it is heterogeneous because it comprises of small and large particle existing together. Heterogeneity arises from the different sizes of sand particles.

Qn. 1.18 physical properties of zinc include the color (silver-gray), the melting point (420 ⁰C). Density and hardness is also a physical property. Chemical properties include its reaction with sulfuric acid to yield hydrogen and subsequent dissolution of the metal; the reaction with oxygen to form zinc oxide.

Qn 1.25 a) ⁰C = (5/9)*(⁰F-32). Therefore, converting 62⁰F yields 5/9(62-32). This results to 5/9(30). Therefore it's 16.67⁰C.

b). ⁰F = (9/5)*⁰C+32. In converting 216.7⁰C yields (9/5) 216.7 +32. 9/5(248.7). This will yield 447.66⁰F.

c). ⁰C= 273K. therefore 233⁰C=> K brings 273+233= the answer is 506K.

d). 315K to F requires conversion to 0C. this brings 315-273= 42⁰C. we then convert to ⁰F. 9/5(42+32)...the answer is 133.2⁰F.

e) 2500⁰F to K. we convert to ⁰C. 5/9 (2500-32). 5/9* 2468. This yields 1371.10⁰C.  we then add 273. The answer is 1644.11K.

Qn. 1.28. a) density= mass/volume. The volume of the osmium is 1.5 x1.5x1.5= 3.375 cm^3. density= 76.31g/3.375cm³ = 22.6103g/cm³.

b) mass= density X volume. Therefore 4.51g/cm³ X125.0 cm³ . remember 1cm³= 1mL. the answer is 563.75g (Theodore et al).

c). 0.1500L= 1500mL. the mass of benzene is therefore 0.8787g/mL X 1500mL. the answer is 1318.05g.

Qn 1.42. a) The conversion factor is 1micrometre =0.001mm.

b) The conversion factor is 1ms= 1000ns.

c) The conversion factor is 1 mile = 1.61kilometres

d). one ft³ is equivalent to 28.3168466liters.

Qn 1.50. The volume of the carbon monoxide is 9.0 X 14.5 X 18.8 = 2453.4 ft³ .  1 ft= 0.02832M³. Mass= density X volume. 48 micrograms/M³ X 0.02832 M³. The answer is 1.35936 micrograms.

Qn. 1.52. if we put Kg as a reference point.. we find that 1 kg is equivalent 2.2 lbs and also equivalent to 0. 3348gallons. In the this scenario therefore shows that 5Kg is heavier, then 1 gal of water while the 5-lb bag is the lightest.

Qn. 3.20 a). C₃H₆(g) + 6O2 => 3CO₂  + 3H2O (g).

b). NH₄NO₃ (s) => N₂O (g) + 2H₂O (g)

c). C₅H₆O (l) + ¹³/₂ O2 (g) => 5CO₂ (g) + 3H₂O (g)

d). N₂ (g) + 3H₂(g) => 2NH₃ (g)

e).  K₂O (s) + H₂0 (l) => 2KOH (aq)

3.22 a). N2O = 14 X2 + 16 = 44g

b) HC₇H₅O₂= 1+ 12 x 7 + 1 x 5 + 16 x 2 = 122g.

c). Mg(OH)₂ = 24 + 2(16+1) = 58g

d). (NH₂)₂CO = 2(14+2) + 12 + 16= 84g

e). CH₃CO₂C₅H₁₁ = 12+1x3+12+ 16x2 +12x5 1x11= 130g

Qn. 3.23 d). % mass of oxygen in C₂₂H₂₄N₂O₈=  128/444 (100%) = 28.828%.

Qn. 3.28 mass of a mole of carbon is 12 grams. A mole of ¹²C contains 6.0221367 x 10²³ atoms.

Qn. 334. a). Mass of CdS is 0.00576 (112.411+28). = 0.8087673grams

b). moles 53.5g/112.6g = 0.475moles

c). 1.305 x 10^-3 X (12x6+6) = 0.1053x 6.0221367 x 10²³ = 6.3413 X 10²² molecules.

d). 0.00488 X (27+ 3(14 +48) = 1.03944moles.

Qn. 3.33b). mole of NH+ in 8.776g of NH₄ CO₃ . (14+4) / (14+4+12+48) = 23.08%. 0.2308 x 8.776g =>>>> 2.025moles .

Qn. 3.38 a). molar mass of aspartame C₁₄H₁₈O₅= 12x 14 +1 x18 + 16 x5= 314grams

b). 0.001g/314g = 3.185 x10^-6 moles.

Qn. 344 a). Reacting 0.104mol of K, with 0.052mol of C, and 0.156mol of O. the ratio of reaction is 0.104/0.052 moles of K: 0.052/052 of C: 0.156/0.052 of O. the ratio is 2:1:3  the emphirical formula is therefore K₂CO₃.

Qn. 348a). (HCO₂)_n = 90 g/mol . (1+12+ 16x2)_n = 90 g/mol. =>>>> (45)_n = 90 . _n = 90/45 =2 Empirical formula =>>>> H₂C₂O_{4 (Theodore et al)}.

Qn. 350c). empirical formula of Epinephrine we find the corresponding ratios of the participating elements in relation to the one with the lowest %. Therefore, 59.0/7.1 of C: 7.1/7.1 of H : 26.6/7.1 of O: and 7.7/7.1of N. this yields 8:1:4:1. The molecular formula is  C₈HO₄N. The empirical formula is arrived at by (C₈HO₄N)_n = 180 amu. (175)_n = 180 . the ratio is 1. The empirical formula is C₈HO₄N.

Qn. 3.58. a). moles of glucose. We can see from the equation that a mol of glucose yields two moles of carbon dioxide.  Therefore, moles of carbon dioxide  produced= 2 x 0.400/1 . this will yield 0.88 moles of Carbon dioxide (Theodore et al).

b). Grams of glucose needed: we can see from the equation that (12x6+12+6x16)g of glucose yields  (2X12+5+16+1) of C₂H₅OH. 180g of glucose yields 46g of carbon dioxide. Grams of glucose needed therefore are > 7.5 x180/46. The answer is 29.35grams of glucose needed.

c) When 44grams of carbon dioxide are yielded in this reaction, 46grams of ethanol is produced. Grams of carbon dioxide produced when 7.5 grams are ethanol is produced are arrived at by finding it as 29.35x44/46. The answer is 28.07 grams carbon dioxide produced.

Qn. 3.72. Aluminum hydroxide is the limiting reactant because its 0.5 moles is equivalent to 61grams reacting. The reacting weight of sulfuric acid on the other hand is 0.5x3x98= 147g. it is observable that sulfuric acid is in excess thus cannot be a limiting reagent. During the reaction, 61g x2 of aluminum sulfate is formed. This equates to 0.35moles

Qn. 5.2. a). This process represents a decrease in internal energy because energy has been lost to products (Theodore et al).

b). loss of energy of system is represented by a negative sign. i.e. -delta E.

c). A process that work does not accompany it is known as endothermic reaction.

Qn. 5.4 Distance is not a state function because it does describe the state of the current system. Elevation on the other hand is a state function because it describes the status of the new system in relation to the initial place.

Qn. 5.12. The kinetic of a tennis ball tossed upwards decreases because the speed of the ball reduces with time, t and height, h.

b). The potential energy is inverse to kinetic energy. It is found that the potential energy will increase to a point where the moving ball will come to rest. This is where the potential will be high.

c). the higher the mass, the sooner the ball will lose the kinetic energy and acquire the potential energy. This therefore means that the height covered will be relatively shorter than the first tennis ball.

Qn. 5.24. a) The first law of thermodynamics deals with the conservation of energy.  Therefore, Change internal energy= Heat added to the system - work done by the system.

b) During an exothermic reaction, heat is lost from the system and this will make the value of Q and W to be negative. If it will be endothermic, the values will be positive.

Qn. 5.26. a). change in energy is 850-382 = + 468J. This reaction is endothermic.

b). Q=mcdT. 50g x4.184J/gK x15 = 3138J. Change in internal energy is 3138-3140= -2J. This reaction is exothermic.

c). the reaction is endothermic since energy in the system increases. That is +6.47kJ.

Qn. 5.29. a state function is a property that is used to describe the equilibrium state that has been acquired by a system irrespective of the path taken to arrive at that state.

b). Examples include location, elevation, longitude, enthalpy and entropy. Work and distance are examples of non-state functions.

c). Work is not a state function because arriving at its value depends on the fact that one has to follow a specific path between the two states of equilibrium. S

qn. 5.32. a). Enthalpy change of a system will be equal to the amount of heat transferred when no work is done by the system.

b). absorption of heat by a system results in the transition of an enthalpy state. When heat is lost to the surrounding, there is drop of enthalpy of a system.

Qn. 5.35. dH= dE+ PdV. Therefore, dE= 18kJ while dH= 61kJ.

Qn. 5.38. a). ZnCO₃ (s) =è ZnO (s) + CO₂ (g)

Qn 5.40. a). 1 mol of CO₂ solid has a higher enthalpy.

b). 2 moles of hydrogen atoms has a higher enthalpy

c). 1 mol of H₂O (g) at 25⁰C. has a higher enthalpy

d). a mol of N2 gas at 300 ⁰C has a higher enthalpy (Theodore et al).

Qn. 5.40. a)A positive value indicates that heat has been absorbed into the system.

b). CH3OH (g)  is equivalent to 32 g. 32g yields +90.7kJ. in order to find the heat released by 45g of ethanol = 45x 90.7/ 32= 127.5kJ.

C). when 90.7 kJ is absorbed, 4grams of hydrogen. To find the grams released when 25.8kJ is released, 25.8 x 4 / 90/7 == 1.14 grams of hydrogen yielded.

Qn 5.46. a).  enthalpy change of the reverse reaction is -630kJ.

b). formation of one mole of acetylene= 630/3=> +210kJ/mol

c). the forward is thermodynamically favored because lots of heat is consumed by the system.

d). the magnitude of the system will decrease because less heat will be absorbed in order to form the gaseous state of acetylene.

Qn. 5.51. Q= mcdT. 1500x 4.184x63.5 ==398.526kJ will be required to raise the temperature.

Qn 5.55. q_qui= mC%u2206T = -qcold = -(mC%u2206T + m%u2206H)

7.13 x 7.854= the answer is 59.999kJ.

Qn. 5.60. a). Enthalpy for the reaction +30  + + 60 kJ = + 90kJ.

Qn. 5.64. N2O (g) +NO2 =è 3NO(g).

3/2(-180.7)- ½(+163.2)- ½ (+113.1). The enthalpy change is therefore, 409.2 kJ.

Qn. 5.65 a). Standard conditions are conditions that are assumed to be stable and are exploited during experimental calculations.

b). enthalpy of formation is the amount energy that is absorbed during the formation of a certain compound.

c) Standard enthalpy of formation is the energy change that occurs due to the formation  of one mole of substance at standard states.

Qn. 5.72. a). the appendix is not accessible.

 

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