385 J of heat are needed to heat a piece of aluminum from 22°C to 45°C. If the specific heat of Al is 0.90 J/g°C, what is the ma
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Answer:
49.35 mL
Explanation:
Given: 56.2 mL of gas
To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg
Solution:
At 820 mm of Hg, volume of gas is 56.2 mL
At 1 mm of Hg, volume of gas is
At 720 mm of Hg, volume of gas is
3.84 Kg of sodium chloride is needed to lower the freezing point of the water so that ice melts.
Explanation:
Data given:
mass of ice = 30 Kg
molar mass of NaCl = 58.5 grams/mole
ΔT(depression in freezing point) = 15.4 F or - 9.22 degrees
Kf = 1.86 degree C/ mole
Molality = ?
mass of NaCl =?
i = 2 (because NaCl breaks into 2 ions)
ΔT = iKfm equation 1
Depression in freezing point when ice melts, the temperature is 0 degrees so
the depression in temperature = 9.22 degrees
putting the values in the equation:
9.22 = 2 x 1.86 x x
mass = 3.84 kg of NaCl will be required to lower the freezing point of the water so that the ice melts.