I must assume that the ice block is at an initial temperature of 0o C. I also need to work in joules,. so I will convert the calories to joules.
1600 cal. (4.184 joule/1 cal.)
= 6694.4 joules
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6694.4 J = (20 grams H2O)(4.180 J/gC)(Tf - 0o C)
6694.4/83.6 = Tf
= 80o Celsius
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The temperature would be that of water's boilng point od 100 degrees
Grams are mass. There's no calculation involved.
Multiply the moles given, by the atomic mass of chlorine and you have your answer. (3.55 Grams)
This question needs additional information to enable me correctly answer it.
Solubility increases with temperature, but the solubility of sodium chloride in water is 316 grams per litre at 0 degrees Celsius, and 330 grams per litre at 70 degrees Celsius. Since room temperature is somewhere between these two, this gives upper and lower limits of the solubility at room temperature. 50 grams of water has a volume of 50 cubic centimetres, or 0.05 litres. In one litre you could dissolve between 316 grams and 330 grams, so in 0.05 litres you could dissolve between 15.8 and 16.5 grams, where 15.8 = 316 x 0.05 and 16.5 = 330 x 0.05. So we can say it's around 16 grams of NaCl in 50 grams of water at room temperature.
The temperature would be that of water's boilng point od 100 degrees
21
The volume depends on the temperature and pressure - neither of which are given.
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
I'll assume here that by "70 temperature" you mean "70 degrees Celsius". Basically, you have to calculate the average temperature of all of the water in the mixture, which will be the final temperature once it's well stirred. The 200 grams of water at 10 degrees represent 2/3 of the total amount of water (300 grams), so thus, multiply 10 by 2/3 to determine their contribution to the final temperature. You will get 20/3. The 100 grams of water at 70 degrees represent 1/3 of the total amount of water, so multiply 70 by 1/3 to determine their contribution to the final temperature. You will get 70/3. When you add together the two temperatures you get 90/3, which is equal to 30. Therefore, the final temperature is 30 degrees Celsius.
It is not possible to answer the question in which the volume of the block is given as Cm Cm and Cm. Some numbers might have been useful!
As the volume of a given gas sample is dependent on its temperature and pressure; to find a volume of any gas which does exist, the temperature and the pressure of the system/vessel should be given directly or could be calculated.
About 232 grams
well you have to think you would weigh this using grams so the density of a foam block is "Grams per cubic centimeters"
875J/12.6g*4.18Jg-1°C+22.9°C=39.5°C
The solubility is expressed in grams of solute dissolved in 1 liter of solvent, at a given temperature.
The solubility is expressed in grams of solute dissolved in 1 liter of solvent, at a given temperature.