Let's use calorimetry to measure heats of reaction but this time let's have a limiting reactant. Review #3 in T8VE if you want. Suppose you mix 25.0 mL of 0.12M cesium hydroxide with 25.0 mL of 0.30M hydrochloric acid. Assume both solutions are at 28.0°C. The temperature of the solution after the reaction is 31.3°C. Because the temperature of the solution went up, the reaction is (endothermic/exothermic). Compute the heat of the reaction assuming the heat capacities of the solutions are equal to pure water (4.18 J/g°C). Report the Q as positive it you thought the reaction was endothermic and negative if you thought it was exothermic. The Q is The heat of the reaction is generally reported as heat per mole. The moles of acid are Look at your moles and pick the limiting reactant. Therefore the heat of the reaction (in kJ/mole) is Joules. and the moles of base are Subm

Let's use calorimetry to measure heats of reaction but this time let's have a limiting reactant. Review #3 in T8VE if you want. Suppose you mix 25.0 mL of 0.12M cesium hydroxide with 25.0 mL of 0.30M hydrochloric acid. Assume both solutions are at 28.0°C. The temperature of the solution after the reaction is 31.3°C. Because the temperature of the solution went up, the reaction is (endothermic/exothermic). Compute the heat of the reaction assuming the heat capacities of the solutions are equal to pure water (4.18 J/g°C). Report the Q as positive it you thought the reaction was endothermic and negative if you thought it was exothermic. The Q is The heat of the reaction is generally reported as heat per mole. The moles of acid are Look at your moles and pick the limiting reactant. Therefore the heat of the reaction (in kJ/mole) is Joules. and the moles of base are Subm

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...

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A student followed the procedure in this experiment, using 29.5 mL of 1.66 molar HCl and 31.5 mL of 1.66 molar NaOH, with both solutions having an initial temperature of 22.1 oC. After mixing, a maximum temperature of 30.7 oC is measured. The final mass of the mixture was found to be 63.194 grams. For this problem, assume the specific heat of the solution is 4.18 J/g oC. What is the heat transferred by this reaction, in Joules? Give your answer in in decimal format.
Could someone please answer this I need to check my answer.
A student followed the procedure in this experiment, using 29.5 mL of 1.59 molar HCl and 31.0 mL of 1.59 molar NaOH, with both solutions having an initial temperature of 22.6 oC. After mixing, a maximum temperature of 33.1 oC is measured. The final mass of the mixture was found to be 64.655 grams. For this problem, assume the specific heat of the solution is 4.18 J/g oC. What is the heat transferred by this reaction, in Joules? Give your answer in in decimal format. When entering units, use proper abbreviated units with proper capitalization.
When 6.30 g sample of solid potassium hydroxide (MM= 56.11g/mol) is dissolved in 101.00 g of water (at 21.2 degrees Celsius) the temperature of the resulting solution is 38.8 degrees Celsius. What is the AH for this dissolution reaction? Assume the specific heat of the solution is the same as water. Only include the numerical value not units or labels.
Please help with questions 5-12. Note: mass of HCl solution = 50.0g and mass of NaOH solution = 50.0g
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Chemistry 14) A 60.0 mL quantity of 0.80 M HCI is mixed with 60.0 mL of 0.80 M KOH in a constant pressure calorimeter with negligible heat capacity. If the initial temperatures of both solutions are equal to 20.24°C. Calculate the heat change of the neutralization reaction if the final temperature of the mixture is 30.0°C. Assume that the densities and specific heats of the solutions are the same as those of water (1.00 g/mL and 4.184 J/g °C, respectively). Note: this point is worth 10 points. A) + 4900.3 J B)-1171.2 J C)-2450.15 J D)-4900.3 J
3
The answer is 44.8 kJ/mol. Show the solution how to get 44.8 kJ/mol.
-2 The equilibrium constant, K, for the following reaction is 1.71x10 at 1120 K 25O₂(g) 250₂(g) + O₂(9) Calculate K, at this temperature for the following reaction SO₂(g) SO₂(g) + 1/2O₂(g)
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Please don't provide handwritten solution ..