Question

please find all the missing blanks.

Data and Calculations: Heat Effects and Calorimetry A. Specific Heat Trial 1 Mass of stoppered test tube plus metal Mass of test tube and stopper 34.564 25.7348 9.150 Trial 2 34.5648 25.7346 एयर 9. ISO 49.6928 Mass of calorimeter Mass of calorimeter and water 48.940 Mass of water Mass of metal 23.6 23.6 Initial temperature of water in calorimeter °C c Initial temperature of metal (assume 100°C unless directed to do otherwise) 101.8 °C 101.8. 0 Equilibrium temperature of metal and water in calorimeter 26.0 25.9 Arwater (Final - Pinitial 9 Specific heat of the metal (Eq. 3) Jigºc J/g °C Approximate molar mass of metal 102T Unknown metal # B. Heat of Solution Mass of calorimeter plus water 38. 135 40.391 Mass of beaker (continued on following page) custom page

Answer 1

How to calculate mass of water?

The amount of heat of a substance can be calculated by:

q = mc delta T

Here, q denotes the heat quantity

m denotes the mass of a substance

c denotes the specific heat capacity

delta T denotes the temperature change

C = water's specific heat capacity = 4.184 joules per gram degree Celsius

delta T = Final - initial temperature

= 40.1 - 22.9 = 17.2

Q = 4.18 joules/g degree celsius

m (H2O)= q/c delta T

= 4.18/4.18 x 17.2

= 0.581 g

Similarly, calculate for metal.

delta T (H2O)= Final - initial temperature

= 40.1 - 22.9 = 17.2

Similarly, calculate for metal.

The molar mass of water = atomic weight of hydrogen x 2 + atomic weight of oxygen

= 1.008 x 2 + 16

= 18.016 g/mol

Similarly calculate for solid.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) + heat of neutralization

= (0.05 L of HCl + 0.05 grams of NaOH)(4.18 J/g °C)(28.6°C - 22.8 °C + 28.6 -21.7/2)

= .0025 x 4.18 x 6.35

= 0.066357 J

Q H2) for reaction = 28.6°C - 22.8 °C + 28.6 -21.7/2

Similarly solve for the remaining part.