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AlexFokin [52]
3 years ago
12

31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams

of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?
Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g). (2 points)
Chemistry
1 answer:
Alexxandr [17]3 years ago
5 0
Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate things. We do as follows:

</span>Heat gained = Heat lost
mC(T2-T1) = - mC(T2-T1) 
31.5C (102.4 - 32.5) = 103.5(4.18)(32.5 - 24.5)
C = 1.57 J/C-g

Hope this helps.
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cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

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4 years ago
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Opal is a hydrated form of silica. If a laboratory analysis of a sample of opal reveals it to contain29.2% Si, 33.3%O, and 37.5%
ZanzabumX [31]
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Explanation:

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