calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30�c, such that the final temperature of the mixture is 5�c. take specific heat capacity of material of vessel as 0.4 j g1�c-1 ; specific latent heat of fusion of ice = 336 j g-1; specific heat capacity of water = 4.2 j g-1�c-1. - chemistry question
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 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) 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… the formulas for converting between degree celsius and degree fahrenheit are: °f = (°c * 9/5) + 32 °c = (°f - 32) * 5/9 to find the temperature when both are equal, we use an old algebra trick and just set of = oc and solve one of the equations. °c = (°c * 9/5) + 32 °c - (°c * 9/5) = 32 - 4/5 * °c = 32 °c = - 32 * 5/4 °c = - 40 °f = (°f * 9/5) + 32 °f - (°f * 9/5) = 32 - 4/5 * °f = 32 °f = - 32 * 5/4 °f = - 40 so the temperature when both the celsius and fahrenheit scales are the same is - 40 degrees. we’re being asked to convert the given temperature from celsius to fahrenheit and kelvin. recall that the conversion from celsius to fahrenheit is given by:°f=1.8(°c)+32[readmore]converting 54 °c  to fahrenheit:°f=1.8(°c)+32°f=1.8(54)+32 of = 129.2 ofrecall that the conversion from celsius to kelvin is given by:k=°c+273.15converting 54 ˚c to kelvin:k=54°c+273.15k = 327.15 k54 °c is equivalent to 129.2 of and 327.15 k 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 [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 the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m