Chemistry, 11.11.2020 23:20 khikhi1705

7.46 - A 0.0500-mol sample of a nutrient substance is burned in a bomb calorimeter containing 2.00 x 10g H20. If the formula weight of this nutrient substance is 114 g/mol, what is the fuel value (in nutritional Cal) if the temperature of the water increased 5.70C?

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Answer from: ccory0626

$200\frac{kcal}{g}$

Explanation:

Hello!

In this case, for this calorimetry problem, since the combustion 0.0500 mol of the nutrient increase the temperature of water by 5.70 °C, we can notice that the heat lost by the nutrient is gained by water in order to write:

$Q_{nutrient}=-Q_{water}$

Which can be also written as:

$Q_{nutrient}=-m_{water}C_{water}\Delta T_{water}\\\\Q_{nutrient}=-200 g*4.184\frac{J}{g\°C}*5.70\°C\\\\Q_{nutrient}=4769.8 J$

Thus, in terms of the grams of the nutrient:

$m_{nutrient}=0.0500mol*\frac{114g}{1mol}=5.70g$

The fuel value in nutritional Cal (kcal/g) turns out:

$Fuel \ Value=\frac{-4.7698kJ}{5.70g}*\frac{1kcal}{4.184kJ}=200\frac{kcal}{g}$

Best regards!

Answer from: Quest

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.