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How many calories of heat are released as 200.0g of water cool from a liquid at 25.0C to ice at -15.0C?
According to the heat equations: Q = C1mDelta;T1 + C2mDelta;T2 + lambda;m. m = 0.2 kg For water: C1 = 4.183x103 [J / (kg K)] Delta;T1 = 25 - 0...
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50.0 g of ice was initially at −4.0°C and 1 atm. At 1 atm, it absorbed some heat and became 20.0 g of ice and 30.0 g of liquid water in equilibrium with each other. How much heat was absorbed?
Let’s first consider the processes that occurred with the initial ice. Firstly, the ice was heated to 0°C (melting point of water). Then, the part of ice melted to form...
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Calculate the heat change of calories to condense 10.0 g of steam at 100 oC, and determine whether heat is absorbed or released
Enthalpy of vaporization for water is 2260 kJ/kg. Thus, the amount of RELEASED heat will be Q = r*m = 2260 kJ/kg*0.010 kg = 22.6 kJ or 5.4 kcal.
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Calculate the heat required to melt 7.87 g of benzene at its normal melting point.
Heat of fusion (benzene) = 9.92 kJ/mol
Mollar mass (C6H6) = 78 g/mol n = 7.87 / 78 = 0.101 (mol) Q = 0.101 * 9.92 = 1.002 (kJ)
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The water of mass 75g at 100°C is added to ice of mass 20g at -15°C.What is the resulting temperature.(Latent heat of ice=80cal/g and specific heat of ice=0.5)approx?
Find the total heat used by ice in changing state from solid to liquid(water) Q1 = 20g*80cal/g = 1600 cal 1600 cal = 6694.4 J = 6.69 kJ M'C(T'-x) =...
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Calculate the energy required to raise 9.8 grams of ice from -10 degrees Celcius to 0 degrees Celcius. Report your answer in Joules. The specific heat of ice is 2.1 j/g/C.
Q=mc∆T.Q= 9.8 x 2.1 x 10 = 205.8 J.
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Calculate the heat needed at 0∘C∘to make each of the following changes of state.
Calories to melt 22 g of ice
QUESTION # 132550Calculate the heat needed at 0∘C∘to make each of the following changes of state.Calories to melt 22 g of iceANSWERSΔHf = q/mWhere;ΔHf = enthalpy of fusion of water(80cal/gram)q...
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Calculate the heat needed (in calories) at 100.0º C to boil 155 g of water.
Heat of water vaporization 2260 kJ/kg. 1 kJ = 0.239 kcal, 2260 kJ/kg = 540.14 kcalQ = 540.14*0.155 = 83.7217 kcal = 83721.7 cal.
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All of the following can be separated by sublimation except: A. Dry Ice and Iodine B. Ammonium chloride and Sodium chloride C. Iron Fillings and Iodine D. Ice cubes and dry ice
Solution :Sublimation is the transition of a substance directly from the solid to the gas phase without passing through the intermediate liquid phase.Dry ice sublimes easily. Iodine will sublimes easily...
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You need to heat 7.59 g of water from 5.2 ºC to 368 K. How much heat is needed? Give answer in Joules and calories. _____J _____cal
5.2°C = (273 + 5.2)k = 278.2kChange in temperature = 368k -278.2= 89.8kq (heat energy) = MCp∆TWhen Cp of water =1046J q = 7.59 g × 1046J × 89.8k =...
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