Can you help me with the numbers 1, 2, 3, 4, 5, and 6 questions? Calculations for Temperature and Phase Change WorksheetThe heat of fusion of ice is 79.7 cal/g.The heat of vaporization of water is 540 cal/g.Report the answer using the correct number of significant figures!1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0gHeat of fusion of ice:100,0 g 79.7cal-1970 cal2. How much energy is required to vaporize 234.5 g of water?234.5g | 540 cal |-30.6 cal2591.2241.2 cal/g1266301 1.2663X105(1.3 x 10³ calAnswer: 1.3 x 105 cal3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization ofthat substance?mass of substance: 25gHeatvaporization: 30.6 cal-500.0g (1 cal) (1.1008) =до- 550 Ca1/4.184579.7cal/gAnswer: 7970 calMass of 234.59Heat of vaporization: 540 cal/gAnswer: 1.2 cal/g4. How much energy is removed from 500.0 g of water when the temperature is lowered by 1.10 °C?-F550 calmass m = 500.09AT= 1.10°C1000.09 1Cal (3x) = 3.00×10²cal3%Q==ssocalAnswer: -550. cal (or -2.30 x 10³ J)23.01.2-2,3012×10³-2.30× 10³5Icg|5. How much energy is required to raise the temperature of 1000.0 g of water from 23.00 °C to26.00 °C?m=1000.09Q=S.M.ATS=1 cal/goe Teinal - TinitialT₁ = 23.00°C12=26.00°CAnswer: 3.00 x 103 cal (or 1.26 x 10¹J)1.26×10 J26.00-23.00=3%8° 3.00x 10³ cal 4 1845 11.2.5 52x104I cal6. The specific heat of copper is (0.0924 cal/g°C), how much energy is required to raise thetemperature of 10.0 g of copper by 100.0 °C?0.0924 cal10.09(357) (100.086)92.4 callmass of copper: 109Specific heat copper: 0.0924cal/g °CAnswer: 92.4 cal

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The heat of fusion of ice is 79.7 cal/g. The heat of vaporization of water is 540 cal/g. Report the answer using the correct number of significant figures! 1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0g Heat of fusion of ice: 100,0g|79.7 call-(1970 cal) 2. How much energy is required to vaporize 234.5 g of water? 234.5g | 540cal |= :79.7 cal/g Answer: 7970 cal Mass of 234.59 Heat of vaporization & 540 callg 12 6630, 1.2663X105 (1.3 x 10³ cal Answer: 1.3 x 105 cal 3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of that substance? mass of substance: 25g Heat vaporization: 30

The heat of fusion of ice is 79.7 cal/g. The heat of vaporization of water is 540 cal/g. Report the answer using the correct number of significant figures! 1. How much energy is required to melt 100.0 grams of ice? Mass of ice = 100.0g Heat of fusion of ice: 100,0g|79.7 call-(1970 cal) 2. How much energy is required to vaporize 234.5 g of water? 234.5g | 540cal |= :79.7 cal/g Answer: 7970 cal Mass of 234.59 Heat of vaporization & 540 callg 12 6630, 1.2663X105 (1.3 x 10³ cal Answer: 1.3 x 105 cal 3. If 30.6 calories are required to vaporize 25 g of a substance, what is the heat of vaporization of that substance? mass of substance: 25g Heat vaporization: 30

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter15: Gases,liquids, And Solids

Section: Chapter Questions

Problem 86E

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Similar questions

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