Answer to Question #49833 in Inorganic Chemistry for Ana Caos




 A sample of gas is heated from -15.0 0C to 28.5 0C. What was its original

volume if the volume at 18.5 0C is 2.65 liters?
The answer to the question is available in the PDF file https://www.assignmentexpert.com/homework-answers/chemistry-answer-49833.pdf
your cubic meter box contains about 1.2 kg of air. the specific heat of air is about 1 kilojoule per kilogram per degree centigrade.

so 1.2 kilojoules per degree centigrade. to get from 20c to 1000c you need 980 * 1.2 = 1176 kilojoules = 327 watt-hours. you can get 1440 watts from a us 15a hous… radius = 1.75 m
height (h) = 12 m
volume = 
= ×  × 1.75 × 1.75 × 12
= 38.5 m3
since, 1 m3 = 1 kiloliter
capacity of the pit = (38.5 × 1)
= 38.5 kilolitres -2.8oc heat lost by the vessel and water= [150 x 4.2 x (30 - 5) j + 100 x 0.4 x (30-5)] j = (150 x 4.2 x 25 + 100 x 0.4 x 25) j = (15,750 + 1000) j= 16,750 j heat gained by ice to change into water at 0 °c = m x specific latent heat of fusion of ice = {m x 336} heat gained by water to come to final temperature (5 °c) = m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m if there is no loss of heat to the surroundings, according to the principle of calorimetry: heat gained by ice to melt and come to final temperature = heat lost by vessel and water or 336m + 21m= 16,750 or 357m = 16,750 or m = 16750/357 = 46.92 setup the kc expression for the reaction, recall that kc expression will appear as:kc expression for the problem:solving for [n2]:note that ice table is not necessary for this calculation answer is 0.016 m mass (molar mass) → mol (avogadro's number) → molecules[readmore]3.51×104 g ×1 mol163.94 g×6.022×1023 molecules1 mol = 1.29x1026 moleculesthere are 1.29x1026 molecules in the sample kclo3(s) — heat→ kcl (s) + o2(g)      reactant        productk        1                   1cl       1                   1o       3                    2[readmore]2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g)      reactant        productk        2                   2cl       2                   2o       6                    6the balanced equation for the reaction is 2 kclo3(s) — heat→ 2 kcl (s) + 3 o2(g) [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v 0.10m hc2h3o2 and 0.10m lic2h3o2 given radius of cone (r) = 3.5/2 m height of cone (h) = 12 m find out we have to find the capacity in kiloleters solution volume of cone = (1/3) × πr2h on taking π=22/7 = (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 = (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2) = 38.5 m3 capacity of conical pit in kilo liters: 1000 m3 = 1 liter 38.5 m3 = 1000 × 38.5 liters = 38,500 liters = 38.5 kilo liters answer the capacity of the container is 38.5 kilo liters