How many molecules of dinitrogen tetroxide (N204) are there in 59 g of N204?
Dinitrogen tetroxide decomposes to produce nitrogen dioxide: N204 (9) - 2 NO2 (g) Calculate the equilibrium constant for the reaction given the equilibrium concentrations at 100°C: [N204] = 0.800 M and [NO2] = 0.400 M A. 5.00 OB.0.200 OC. 0.500 OD.2.00
Question 7 At low temperature nitrogen dioxide molecules join together to form dinitrogen tetroxide. 2 NO2(g) + N204(9) (low temperature) A sample of NO2 sealed inside a glass bulb at 23 °C gave a pressure of 673 Torr. Lowering the temperature to -5 °C converted the NO2 to N204. What was the final pressure (in Torr) inside the bulb? Torr
5.00 L flask is evacuated and 47.65 g of solid dinitrogen tetroxide, N204, is introduced at-196 ℃. The sample is then warmed to 25 °C during which time the N204 vaporizes and some of it dissociates to form brown NO2 gas. The pressure slowly increases until it stabilizes at 3.51 atm. a. Write the balanced reaction for the dissociation of N204 b. What would be the pressure in the flask at 25 °C if the gas were all N204? Number...
2. For the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N204 = 2 NO2 Suppose that both the forward and reverse reactions are elementary processes with rate constants of k, and kr respectively, and the equilibrium constant for the process is 14.48 at 298 K. If the rate of the reverse reaction is 1324 M/s when (NO2) = 0.60 M, what is the half-life for the decomposition of dinitrogen tetroxide (the forward reaction) at the same temperature?
A mixture of dimethylhydrazine (CH3)2N2H2 (MM-60 g/mol) mixed with dinitrogen tetroxide (N204) are very effective rocket fuels. The reaction between the two reagents is give below: (CH3)2N2H2 (1)+2N204() 3N2(g) + 4H20(g) + 2C02(g) If 1.30 x 102 g of dimethylhydrazine reacts with excess N204 and the product gases are collected at 35°C in a 500 L tank. What is the partial pressure of N2 gas produced and what is the total pressure in the tank once the reaction is complete?
1) Dinitrogen tetroxide partially decomposes according to the following equilibrium : N204 (9) = 2'N02 (9). A 1.00 L-flask is charged with 0.0400 mol of N204. At equilibrium clt 373 K, 0.0055 mol of N204 remains. Key for this reaction is -
I NEED HELP WITH THIS PROBLEM SOMEONE HELP PLEASE!!! Dinitrogen tetroxide partially decomposes according to the following equilibrium: N204(g) → 2NO2(g) A 1.000-L flask is charged with 2.60 x 10-2 mol of N204. At equilibrium, 2.01 x 10 2 mol of N204 remains. Keg for this reaction is O 0.361 O 1.53 x 10-4 O 0.615 6.94 x 10-3
QUESTION 22 The decomposition of dinitrogen tetroxide to nitrogen dioxide at 400°C follows first-order kinetics with a rate constant of 9.17 X10sl. Starting with pure N204, how many minutes will it take for 85.0% to decompose? *Please report 3 significant figures. Numbers only, No unit. No scientific notation.
Part 2: Mole Calculations (8 pts) 1. How many moles of sodium carbonate are there in 75 grams of K2COs? 2. How many moles of each atom are there in 5 moles of Cus(PO)2 3. How many molecules of dinitrogen tetroxide (N204) are there in 80 g of N204? 4. How many grams does one molecule of carbon dioxide (CO2) weigh? The balanced equation for production of carbon dioxide is C(s) + Oz(g) → CO2(g). How many grams of CO2are...
The reversible decomposition of dinitrogen tetroxide, N, O, to nitrogen dioxide, NO,, is shown. For this reaction, Keq = 0.15. = NO dinitrogen tetroxide 2NO, nitrogen dioxide At equilibrium, is the concentration of reactants or products greater? The concentration of reactants equals the concentration of products at equilibrium. The concentration of products is greater than the concentration of reactants at equilibrium. The concentration of reactants is greater than the concentration of products at equilibrium Consider this system at equilibrium. PC12(g)...