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BlackZzzverrR [31]
2 years ago
6

When a sample of aqueous hydrochloric acid was neutralized with aqueous sodium hydroxide in a calorimeter, the temperature of 10

0.0 g of water surrounding the reaction increased from 25.0°c to 31.5°c. if the specific heat of water is 1.00 cal/(g•°c), calculate the quantity of energy in calories involved in this neutralization reaction?
Chemistry
1 answer:
Annette [7]2 years ago
4 0
<span>6.50x10^3 calories. Now we have 4 pieces of data and want a single result. The data is: Mass: 100.0 g Starting temperature: 25.0°C Ending temperature: 31.5°C Specific heat: 1.00 cal/(g*°C) And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this: Let's start with specific heat with the unit "cal/(g*°C)". The cal is what we want, but we'ld like to get rid of the "/(g*°C)" part. So let's multiply by the mass: 1.00 cal/(g*°C) * 100.0 g = 100.0 cal/°C We now have a simpler unit of "cal/°C", so we're getting closer. Just need to cancel out the "/°C" part, which we can do with a multiplication. But we have 2 pieces of data using "°C". We can't multiply both of them, that would give us "cal*°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So 31.5°C - 25.0°C = 6.5°C So we have a 6.5°C change in temperature. Now let's multiply: 6.5°C * 100.0 cal/°C = 6500.0 cal Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories. Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.</span>
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Mass of carbon dioxide = ?

Mass of oxygen = ?

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First of all we will write the balanced chemical equation,

C₆H₁₂O₆  + 6O₂       →    6CO₂  + 6H₂O

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Now we compare the moles of oxygen with glucose from balance chemical equation.

                             C₆H₁₂O₆          :              O₂  

                                    1                :              6

                                    0.322       :              0.322×6 = 1.932 mol

Mass of oxygen:

Mass of oxygen = number of moles × molar mass

Mass of oxygen =  1.932 mol × 32 g/mol

Mass of oxygen =  61.824 g

Now we compare the moles of carbon dioxide with moles of glucose and oxygen.

                              C₆H₁₂O₆            :              CO₂

                                   1                   :                 6

                                  0.322           :           0.322×6 = 1.932 mol

                                   

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                                  6                     :                  6

                                 1.932                :                  1.932

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mass of carbon dioxide = number of moles × molar mass

mass of carbon dioxide =  1.932 mol × 44 g/mol

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