candiishop

phase 1) q= 1.24 mol x 132J/Kmol x (43-23)K = 3274J = 3.27kJ phase 2) q= 1.24 mol x Delta Hm = 1.24 mol x 2.5kJ mol-1 = 3.1kJ phase 3) q= 1.24 mol x 132J/Kmol x (77+23)k = 16368J = 16.4 kJ phase 4) q= 1.24 mol x delta Hv = 1.24 mol x 33kJmol-1 = 40.92 kJ phase 5) q= 1.24 mol x 83J/Kmol x (109-77)K = 3293.44J = 3.29 kJ :. Total energy (q)= 3.27 + 3.1 + 16.4 + 40.92 + 3.29 = 67.0kJ :. B

Is the answer B?

Hi, I'm having trouble answering this question. I don't know how to start. Any help is greatly appreciated Thanks. Using the data below for CCl4, calculate the total energy required to heat 1.24 mol of CCl4 from -43°C to 109°C. Cp(s) = Cp(l) = 132 JK-1mol-1, Cp(g) = 83 JK-1mol-1 Cp: molar heat capacity melting point = -23°C, boiling point = 77°C Hm (melting) = 2.50 kJmol-1, Hv (Heat vaporisation)= 33.0 kJmol-1 A) 23.0 J B) 67.0 kJ C) 60.3 kJ D) 56.9 kJ E) 22.9 kJ

Which of the following properties of a real gas is related to the 'b' coefficient in the van der Waals' equation? (P + n2a/V2)(V - nb) = nRT a) There are attractive forces between atoms or molecules of a real gas b) The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the gas c) Real gases consist of molecules or atoms which have volume d) The average speed of the molecules of a real gas increases with temperature e) None of the above Do you think the answer is c? Thanks

Another question... (sorry such a clueless in this topic) Predict what would happen to erythrocytes placed in a solution containing 300mOsm/L urea and 300mOsm/L sucrose (i.e. a total of 600mOsm/L of particles in one solution; and remember that this is NOT made by mixing equal volumes of 300mOsm/L urea and 300mOsm/L sucrose). Fully explain your answer. My draft response: As urea can pass through the membrane and the sucrose can't. 150mOsm of the urea will pass into the cell making the cell a total of 450mOsm...

Hi.. I having trouble answering this question. Any input would be greatly appreciated =) Explain why the size of red blood cells stored in 0.15M urea would differ significantly from cells stored in the same concentration 0.15M of Na Cl? (Hint: work out the osmolarity and tonicity of the 2 solutions and compare each solution to the intracellular fluid of the erythrocyte(Red blood cells) This is what I done so far... not sure whether it's right or not. Osmolarity Urea: 0.15 x 1000 = 150mOsM Salt [Na Cl]: 0.15 x 2 (the sodium and the chloride ions dissociate and 2 particles formed) x 1000 = 300 mOsM Tonicity Urea: cell bigger (swell) and becomes normal... isotonic solution Salt: Smaller

In distillation, would increasing the heating rate affect the boiling point? Why? Explain in terms of vapour pressure and the pressure of the laboratory.

Ok... sorry i didn't know this forum was for discussion and I'm currently new to this site. I don't really need answers... i just need some idea how to do this question. I thought this question is confusing as well... but what i written up there is correct. I'm just helping a friend from a different college so it's like a revision for me too. Thanks anyways

67877 L of concentrated nitric acid (74% by mass, density 1.46 g/cm3) was split on a road. Sodium carbonate was spread on the acid to react as in the equation: (CO3)2(aq) + 2H ------> CO2(g) + H2O. How many grams of sodium carbonate are needed? If you answer is in the form 5.00 x 102 write in 5.00e2

Many thanks for reply although i'm still unsure how to do the question or my answers are correct.

The decomposition of sulfuryl chloride, SO2Cl2, is described by the following chemical equation: SO2Cl2(g) <-----> SO2(g) + Cl2(g) (i) At 670 K, the following equilibrium concentrations were observed: [sO2Cl2] = 0.0678 M, [sO2] = 0.0243 M, and [Cl2] = 0.121 M. Calculate the equilibrium constant Kc for the reaction at 670 K. Is this right? Kc = [sO2] [Cl2] = 0.0243 x 0.121 = 0.00294 (ii) From your answer in (i), which side of the reaction does the equilibrium favour at 670 K? Justify your choice. (iii) At the higher temperature of 1250 K, the value of Kc is 85.3. If 0.152 moles of SO2Cl2(g), 0.336 moles of SO2(g), and 2.34 moles of Cl2(g) were placed in a 2.50 L flask at 1250 K, would the amount of SO2 increase or decrease as the reaction proceeded to equilibrium? Show your working. SO2Cl2(g) <-----> SO2(g) + Cl2(g) 85.3 = [x] [x] [x]^2 = 85.3 [x] = square root of 85.3 :. [x] = 9.236 M

I have to design a home experiment and stimulate thinking about hypotheses, variables, controls, replicates, experiments and conclusions on growing microbes. I'm thinking of boiling a strap of grass (nutrient--- all microbes need basic nutrients which contain the elements required to build proteins etc) and placed it in an enclosed jar for several days and bring the samples to the labs for staining and examination. My control for this experiment will be that all jars will have to be the same and cleaned and sterilised in exactly the same manner and I don’t know what I should write for the hypothesis. Any further suggestions? Please help. Most people in my class are doing bread mold but it's so disgusting when testing it! lol

Many thanks for the reply. For the first question, did i do my working out right to show how 1mg could dissolve?

Anyone here did chemical equilibrium? I'm stuck on this question. Any help would be much appreciated. 1) for AgCl has a value of 1.8 x 10-10 at 25 degrees celsius, calculate whether or not 1 mg AgCl could dissolve in 1.0 L water.If Ksp AgCl (s) <----> Ag+ (aq) + Cl- (aq) Ksp = [Ag+] [Cl-] = [x] [x] = [x2] x2= 1.8 x 10^-8 [x] = 1.34x10^-5 mol/l = 143g/mol = 0.0018g/L ?????? 2) Although the solubility of AgCl is extremely low in pure water, it is even lower in a 0.1M solution of sodium chloride (NaCl) due to the ‘common- ion effect’. Briefly explain this observation.