### Educational Transcription**Problem Statement**: How much heat is ______ evolved/absorbed when 2.22 x 10² moles of N₂ are consumed?\[ \text{N}_2 + 3 \text{Br}_2 \rightarrow 2 \text{NBr}_3 + 44.0 \, \text{kJ} \]**Tasks**:1. Determine if the reaction is ______ endothermic/exothermic.2. Calculate heat in ______ kJ.3. Use the equation: \[(2.22 \times 10^2 \, \text{mol N}_2)(\text{----------------------------}) = \, \text{________} \, \text{kJ}\]**Options**:- a. 1- b. 2- c. 3- d. 4- e. 5- f. 253.719- g. \(3.38 \times 10^0\)- h. \(3.26 \times 10^3\)- i. 44.0- j. \(1.33 \times 10^{-2}\)- k. \(9.77 \times 10^3\)- l. \(4.44 \times 10^2\)- m. \(6.66 \times 10^2\)- n. 500.0- o. 556- p. 90.0- q. 3.2852- r. 159.81- s. 28.013- t. 1.750- u. 0.8750- v. 13.0- w. 0.46306- x. 1.3892- y. 24.5- z. 38.5- aa. 2.190- bb. evolved- cc. absorbed- dd. exothermic- ee. endothermic

The given reaction is as follows: N2 + 3Br2 → 2NBr3 + 44.0 kJ The moles of N2 given = 2.22×102 molA reaction is said to be the exothermic reaction if the heat is evolved in the reaction. It is written on the product side with a plus sign. As we can see that in the reaction the heat is evolved so the reaction is an exothermic reaction. The amount of energy evolved is 44.0 kJ. Calculation of heat produced: 2.22×102 mol N244.0 kJ1 mol N2=9768 kJ=9.768×103 kJ=9.77×103 kJ Hence, the amount of energy that evolved is 9.77×103 kJ. How much heat is bb. evolved when 2.22 x 102 moles of N2 are consumed? N2 + 3Br2 → 2NBr3 + 44.0 kJ The reaction is dd. exothermic. 2.22 × 102 mol N2 i. 44.0 kJa.1 mol N2 = k. 9.77 x 103 kJ

Answered: How much heat is N₂ + 3 Br₂ This…

Science

Chemistry

How much heat is N₂ + 3 Br₂ This reaction is 2 NBr3+ 44.0 kJ (2.22 x 102 mol N₂)(- evolved/absorbed when 2.22 x 10² moles of N₂ are consumed? endothermic/exothermic kj -) = mol N₂ kj

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

48.2 mL of a solution of HCl is combined with 110.7 mL of 1.35 M NaOH in a calorimeter. The reaction mixture is initially at 22.4°C, and the final temperature after reaction is 29.3°C. What is the molarity of the HCl solution? You may assume that there is an excess of base (so that all the HCI has reacted), that the specific heat of the reaction mixture is 0.96 cal/g °C, and that the density of the reaction mixture is 1.02 g/mL. The heat of neutralization of HCl and NaOH is 13.6 kcal/mole.
An ideal gas in a cylinder fitted with a piston expands at constant temperature from a pressure of 9 atm and a volume of 4.,0 L to a final volume of 13 L against a constant opposing pressure of 1.4 atm. How much heat does the gas absorb, expressed in units of L atm (liter x atm)? Latm
In a constant-pressure calorimeter, 55.0 mL of 0.350 M Ba(OH), was added to 55.0 mL of 0.700 M HCI. The reaction caused the temperature of the solution to rise from 23.81 °C to 28.58 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g °C.) respectively), what is AH for this reaction (per mole H₂O produced)? Assume that the total volume is the sum of the individual volumes. AH= kJ/mol H₂O
3
A piece of molybdenum metal whose mass is 6.87 g is heated to 100.0 °C and then dropped into an ice calorimeter. As the molybdenum cools to 0.0°C, 0.513 g of ice melts (it takes 6.01 kJ of heat to melt exactly 1 mol of ice). What is the molar heat capacity (in J/mol·K) of molybdenum? i J/mol*K
The reaction below is a exothermic or endothermic reaction. 2SO3(g) + Heat ⇋ 2SO2(g) + O2(g)
F HNO3 (1 mole) H2SO4, heat ?
Consider an endothermic reaction that takes place in 303 mL of solution. If the reaction requires 4,468 J of energy and the initial temperature of the solution is 26.7, what will the final solution temperature be? (3 sf) The density of solution is 1.00 g/mL and the specific heat of the solution is the same as that of water.
A 5.43 g sample of an unknown salt (MM = 116.82 g/mol) is dissolved in 150.00 g water in a coffee cup calorimeter. Before placing the sample in the water, the temperature of the salt and water is 23.72°C. After the salt has completely dissolved, the temperature of the solution is 28.54°C. If -3.13 x 10 J of heat was lost during the dissolution reaction of 0.0465 moles of the unknown salt, what is the enthalpy change (in kJ/mol of salt) for the dissolution reaction? Type here to search
8. Instant cold packs can be made from ammonium nitrate by mixing/reacting it with water (reaction shown below). When 1.25 g of NH4NO3 is dissolved in 25.0 g of water at 25.8 °C, the temperature of the water after the reaction is 21.9 °C. What is the enthalpy change (ΔH) for this reaction in units of kJ/mol of NH4NO3? The heat capacity of water is 4.18 ?/ ?℃ . NH4NO3(s) → NH4+(aq) + ??3 − (aq)
In a constant-pressure calorimeter, 65.0 mL of 0.910 M H₂SO, was added to 65.0 mL of 0.300 M NaOH. The reaction caused the temperature of the solution to rise from 24.00 °C to 26.04 *C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g °C), respectively), what is AH for this reaction (per mole of H₂O produced)? Assume that the total volume is the sum of the individual volumes. AH = V 5 R B 6 MacBook Air N 39 8 M command P option kJ/mol H₂O
When a 0.740-g sample of trinitrotoluene (TNT), C,H,N,O, is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter and its content is 3.35 kJ/°C. How much heat was produced by the combustion of the TNT sample? O 34.2 kJ O 11.7 kJ O 101.5 kJ