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Radda
22 days ago
14

Calculate ΔS°rxn (J/k) for 3NO2(g) + H2O(l)LaTeX: \longrightarrow⟶ NO(g) +2HNO3(l) C6H12O6(s) + 6O2(g) LaTeX: \longrightarrow⟶ 6

H2O(g) +6CO2(g) Enter numbers to 1 decimal places. Substance or Ion S° (J/molLaTeX: \cdot⋅K) N2(g) 191.5 N2O(g) 219.7 NO(g) 210.65 NO2(g) 239.9 F2(g) 202.7 H2(g) 130.6 HNO3(l) 155.6 HNO3(aq) 146 H2O(l) 69.940 H2O(g) 188.72 C6H12O6(s) 212.1 O2(g) 205.0 CO2(g) 213.7 CO2(aq) 121 NF3(g) 260.6
Chemistry
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One hour of bicycle riding can require 500-900 kcal of energy, depending on the speed, the terrain, and the weight of the racer.
alisha [2963]
145 hours. Explanation: Riding a bicycle for one hour expends 505 kcal of energy. Given that one gram of body fat equals 7.70 kcal, and 1 pound of body fat is equivalent to 454 grams: 1 lb = 454 g; thus, 21 lb = 21 × 454/1 = 9534 g. Moreover, converting 9534 g of body fat gives us 9534 × 7.70 kcal/1 = 73411.8 kcal. If riding for one hour burns 505 kcal, then to lose 73411.8 kcal, it would require 73411.8 kcal x 1 hour/505 kcal = 145 hours.
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2 months ago
Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g
Anarel [2989]
An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.
7 0
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