SIZING THINGS UP: For each pair, indicate the LARGER quantity. If the two quantitiesare equal, write "equal", and if there is not enough information to determine which is largerstate "not enough info. Explain your reasoning. YOU MAY NOT USE A CALCULATORFOR THIS PROBLEM. Use your estimation skills!a. A solution's temperature afteran endothermic reaction occursA solution's temperature afteran exothermic reaction occurs(initial temperature = 5°C)(initial temperature = 5'C)%3Db. The final temperature when50.0 g of ice at O°C is addedto 100.0 g of water at 80°C.The final temperature when 50.0g ofwater at 0°C is added to 100.0 g of waterat 80.0°C.C. The specific heat of a materialif 500.0J raises the temperature >of 5.0 grams of it by 10.0°C.The specific heat of a material if 500.0Jraises the temperature of 12 g of itby 5.0°C.

#(a): When a chemical reaction takes place inside a solution, an endothermic reaction absorbs heat from the solution. Hence the temperature of the solution decreases. An exothermic reaction releases heat to the solution. Hence the temperature of the solution increases. So the temperature of the solution decreases in case of an endothermic reaction, and increases in case of an exothermic reaction. Hence a solution's temperature after an exothermic reaction is larger (Answer) #(b): In the case of 50 g of ice at 0 oC, it needs extra heat to melt from ice at 0 oC to water at 0 oC. Hence ice at 0 oC will absorb more heat from 100 g water at 80 oC. Hence the final temperature is smaller in comparison to that of water at 0 oC. Hence the final temperature when 50.0 g of water at 0 oC is added to 100.0 g water at 80.0 oC is larger (Answer) #(c): We have to use the relation Q = m*C*∆T to make the approximation. Since Q (=500.0 J) is constant, specific heat capacity(C) is inversely proportional to the product of m and ∆T. For 5.0 g and 10.0 oC, m*∆T = 5.0*10.0 = 50.0 For 12.0 g and 5.0 oC, m*∆T = 12.0*5.0 = 60.0 Since m*∆T is smaller for 5.0 g and 10.0 oC, its specific heat is larger. Hence the specific heat of material if 500.0 J raises the temperature of 5.0 grams of it by 10.0 oC is larger. (Answer)

Science

Chemistry

SIZING THINGS UP: For each pair, indicate the LARGER quantity. If the two quantities are equal, write "equal", and if there is not enough information to determine which is larger state "not enough info. Explain your reasoning. YOU MAY NOT USE A CALCULATOR FOR THIS PROBLEM. Use your estimation skills! a. A solution's temperature after an endothermic reaction occurs A solution's temperature after an exothermic reaction occurs く (initial temperature = 5°C) (initial temperature = 5°C) %3D %3D b. The final temperature when 50.0 g of ice at O°C is added to 100.0 g of water at 80°C. The final temperature when 50.0g of water at 0°C is added to 100.0 g of water at 80.0°C. C. The specific heat of a material if 500.0J raises the temperature of 5.0 grams of it by 10.0°C. The specific heat of a material if 500.0J raises the temperature of 12 g of it by 5.0°C.

SIZING THINGS UP: For each pair, indicate the LARGER quantity. If the two quantities are equal, write "equal", and if there is not enough information to determine which is larger state "not enough info. Explain your reasoning. YOU MAY NOT USE A CALCULATOR FOR THIS PROBLEM. Use your estimation skills! a. A solution's temperature after an endothermic reaction occurs A solution's temperature after an exothermic reaction occurs く (initial temperature = 5°C) (initial temperature = 5°C) %3D %3D b. The final temperature when 50.0 g of ice at O°C is added to 100.0 g of water at 80°C. The final temperature when 50.0g of water at 0°C is added to 100.0 g of water at 80.0°C. C. The specific heat of a material if 500.0J raises the temperature of 5.0 grams of it by 10.0°C. The specific heat of a material if 500.0J raises the temperature of 12 g of it by 5.0°C.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Use the following information to answer the next three questions Electrowinning of Copper Battery Thin sheet of pure copper (Cathode) Thick block of impure Cu (Anode) - Tank Cu2+ Acidified copper sulphate solution (Electrolyte) Cu2+ Impurities (Anode mud) The cell shown above is used to purify copper. At one electrode, impure copper is converted into copper(II) ions, which are later converted into copper solid of high purity at the other electrode. 6. The plating of pure copper will occur at the cathode where oxidation occurs. O cathode where reduction occ occurs. O anode where reduction occurs. O anode where oxidation occurs. 7. During the operation of this cell O the total energy of the system decreases. O the pH of the electrolyte increases. O the predicted net cell potential is 0.00 V.
Converting Concentrations to Different Units Introduction: Depending on how the concentration unit is defined, concentration can serve as a conversion between the amount of a solute and the amount of a solution or solvent. Concentrations can therefore be helpful in a number of stoichiometry issues. The original definition of the concentration unit is frequently preferred because it contains the conversion factor. Today, for chemists, it is suitable to make a solution in the lab using a specific concentration unit like molarity, and then convert the concentration to another unit. Today in the lab, I will prepare a sodium bicarbonate (baking soda) solution with a specific molarity. I will then translate the concentration into molality, mass percent, and mole fraction. Procedure The first step of this lab was to make sure I had everything I needed and that my surroundings were safe. I then looked at an empty beaker weighing paper and a balance. I first took a piece of weighing…
diagram b is correct, What procedures can be used to solve this problem? write down the plan for this new way to solve the problem. 1. first execute plan 1 to solve this problem. explain in every step. 2. solve the problem again - this time use the plan that you propose, plan 2. 3. finally, think about the way that you can use it to make sure that your answer makes sense. write your assessment.
Question 2/3 please
When making a solution, the container gets hot. This is an example of an endothermic process exothermic process
Q9 chapter 15 preparing solutions Solve Asap please
Please help me with this question
What is the molarity of a solution that is composed of 2.5 moles of NaOH dissolved in enough water to make 1,255 of solution? (Report your answer rounded to two decimal places ) Show all work
When a solution is being diluted, a) more solvent is added so that the concentration of solute decreases. b) more solute is added so that the concentration of solute increases. c) some solute is removed so that the concentration of solute decreases. d) more solute is added so that the concentration of solute increases.
A student conducts an experiment. The student prepared 3 beakers which sequentially contained copper metal, zinc metal and 1M sulfuric acid solution. Which of the following statements is true…. A. Copper will dissolve to produce H2 . gas B. When the two metals are connected by a wire, the copper will dissolve C. Zinc metal will dissolve to produce H2 . gas D. Both metals will dissolve
C Determine the moles of sodium carbonate that need to be taken from the concentrated solution. (To avoid introducing rounding errors on intermediate calculations, enter your answer to four significant figures.) Moles from the concentrated solution = Show/Hide Help Submit mol Concentrated solution conc. Diluted solution conc. Diluted solution volume Diluted solution moles 2.45 M 0.250 M 100.0 mL 0.02500 mol
blue. The reaction produces a reddish brown color copper metal and clear aluminum nitrate solution. Beaker Before Reaction Beaker After Reaction Blue colored solution Clear solution Silver piece of Al Small silver piece of Al Dark reddish brown solid According to the following data, what is the limiting reactant? Copper Nitrate solution, because the solution was blue at the end of the reaction showing some copper nitrate remained at the end of the reaction. Aluminum, because some Aluminum remained at the end of the reaction. Aluminum, because it was completely consumed in the reaction. Copper nitrate solution, because the solution was clear at the end showing it was completely consumed in the reaction.