WebQuestion: A 0.0651-mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 8.35 x 102 g H2O. Given that the fuel value is 4.78 x 101 in nutritional Cal when the temperature of the water is increased by 4.26°C, calculate the fuel value in kJ. Enter your answer in scientific notation. WebAll tutors are evaluated by Course Hero as an expert in their subject area. Answered by sabazaman24 The heat of combustion of hydrocarbon is - 41.04 KJ/mol Step-by-step explanation Data Given: mass of hydrocarbon= 0.0651 mole Raise in temperature= 2.19 Celsius Heat capacity= 1.229 KJ/C To find: heat of combustion=?? Solution:
How many moles of iron(II) bromide, FeBr2, are there in 339 mL of …
Web3. MsAnyaForger. To calculate the number of moles of FeBr2 in the given solution, we can use the formula: moles of solute = concentration x volume First, we need to convert the volume from mL to L: 339 mL = 0.339 L Then, [ we can plug in the values we have: moles of FeBr2 = 0.203 M x 0.339 L moles of FeBr2 = 0.068877 mol Therefore, there are 0. ... WebA 0.0651-mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 8.35 x 102 g H2O. Given that the fuel value is 4.78 x 101 … taxpayer by pan
Solucionario Introduccion A La Fisicoquimica Thomas-engel-y …
WebE1B.6 (b) Evaluate the collision frequency of O2 molecules in a gas at 1.00 atm and 25 °C. 395 pm. Calculate (i) the mean speed of the molecules, (ii) the mean free path, (iii) the collision frequency in air at 1.0 atm and 25 °C. E1B.7 (b) The best laboratory vacuum pump can generate a vacuum of about 1 nTorr. WebJan 15, 1998 · The free reaction enthalpy Delta G degrees for this process is calculated to be 4.7 kcal/mol, which explains the thermal instability of N(2) complex 2a with respect to the loss of N(2). Web14-1 (a) The initial pH of the NH 3 solution will be less than that for the solution containing NaOH. With the first addition of titrant, the pH of the NH 3 solution will decrease rapidly and then level off and become nearly constant throughout the middle part of the titration. taxpayer center