Question

A 500.0-g sample of an element, at 195°C is dropped into an ice-water mixture. 1095 gofioevoelka und an ice-water mixture remains. Calculate the specific locat of the elencia, mod determine which element it is. В. Ва A. Zn C. Po D. Ag 33. What is the final temperature, in °C, when 20.0 g of water at 80°C is mixed with 200 g of water 1292 A. 12°C B. 7.0°C C. 8.8°C D. 525°C (34.) How much ozone is used if the following reaction releases 568 kJ of energy in the form of heal Pb(s) + C(s) + 03(8) - PCO(S) AIP -? B. 1928 A. None D: 488 C. 48 D. 2307 3 (35) C. 4025J/K Benzoic acid, CH.02, is a standard used in determining the heat capacity of a calorimeter. Alt* of combustion of benzoic acid is 3.22 x 10 kJ/mol. 0.5 g of benzoic acid was burned in a calorimeter containing 1000.0 g of water. The change in temperature of the calorimeter was 3°C. Calculate capacity of the calorimeter in J/K. · B. 210J/K. A. 450 J/K . is highly combustible and creates a fire hazard 4*+2016) + SIO2()+24,018) AH?=? loving information:

questions 32,33, and 34. i need help with them asap.

and 35

Answer 1

Q32 :

Heat absorbed by ice = mass× latent heat of fusion of ice

= 109.5g × 333.55J/g = 36523.725J

Heat lost by metal = m.s.∆T = 500.0g× s × 195℃ = 36523.725J

s = 0.3746 J/g.℃

Specific heat of element = 0.375 J/g.℃. (answer)

Atomic mass × specific heat of metal = 3R

Atomic mass = 3×8.314/0.375 =66.58 g/mol

Atomic mass is very close to that of Zn.

Answer: (a) Zn

Q33: heat lost by hot water = heat gained by cold water

m1s∆T1 = m2s∆T2

Let final temp is t℃

20.0g × s × (80-t) = 20.0g×s×(t-25)

t = (80+25)/2 = 52.5℃

Answer: (d) 52.4℃