calculate heating 235 g of water from 22.6c to 94.4c in a | studysoup
calculate heating 235 g of water from 22.6c to 94.4c in a microwave oven requires 7.06 1 0 4 j of energy. if the microwave frequency is 2.88 1 0 10 s -1 , how many quanta are required to supply the 7.06 1 0 4 j?
chapter 5 key: electron: e                         number: # ­                 valence electron: ve   three­dimensional: 3d molecular: molec               with: w/ chemical: chem                 different: diff positive: pos negative: neg i. bonding theories ­ a. valence shell e pair repulsion (vsepr)  valence shell e pair repulsion: model predicting the arrangements of ve  ­ pairs around a central atom that minimizes their mutual repulsion  electron pair geometry: 3d arrangements of bonding pairs andn lone pairs  of e about a central atom  molecular geometry: 3d arrang
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… 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 for water, h2o1 mol h2o = 18.0 g h2o1 mole h2o = 6.022x1023 h2o molecules[readmore]the number of h2o molecules in a 9.00-g sample of water:9.00 g h2o×1 mol h2o18.0 g h2o×6.022×1023 h2o molecules1 mol h2o= 3.01x1023 h2o moleculesthe number of h2o molecules in a 9.00-g sample of water is 3.01x1023 h2o molecules. 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 [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 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 Max 

Hello Max,


There is a little typo on the solution. In a warmer winter you will have gas consumption reduced by 10%, that implies a reduction of 10% of sales and variable costs (the hypothesis is to consider them linearly dependant).


Talink about the Infrastructure costs, they are composed by fixed (60%) and variable(40%) costs. Applying the saving on the variable part you have:


240*40%*10%=9.6  

  

So that the new Infrastructure costs will be 240-9.6=230.4  

  

Does it make sense?  

Feel free to contact me if you have any doubt.  

  

Luca 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 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 Correct option is A) 9.378 m2


![](https://lh3.googleusercontent.com/-UueuqyijXPQ/YLtK0v0m9CI/AAAAAAAABro/xFcfQOQAOx8jHAtpqiORvW_8DRFX1yK3ACLcBGAsYHQ/image.png)


result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits)