the mass of ice taken = 10 g
the mass of water = 250 g
initial temperature of water = 20 C
the final temperature of water = 16. 8 C
specific heat of water = 4.18 J/g*K
the heat absorbed by ice to melt = heat loss by water
heat loss by water = mass X specific heat of water X change in temperature
heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules
heat gained by ice = 3344 J
heat gained by ice = enthalpy of fusion X moles of ice
moles of ice = mass / molar mass = 10 / 18 = 0.56 moles
enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole
Answer:
8.8 yards per second
Explanation:
50 - 20 = 30 yards
30 yards/3.4 seconds = 8.8235
Answer:
0.18 moles
Explanation:
Applying,
PV = nRT................... Equation 1
Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol
Substitute these values into equation 2
n = (0.8316×5.3)/(0.083×295)
n = 0.18 moles
