What is Delta. S for the reaction N2(g) O2(g) Right arrow. 2NO? Use Delta. G = 173. 3 kJ at 303. 0 K and Delta. H = 180. 7 kJ. Use Delta. G = Delta. H – TDelta. S. 7. 4 J/K 7. 4 kJ/K 24. 42 kJ/K 24. 42 J/K.

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

Caso4(s) < > ca+2 + so4^-2 ksp = [ca+2][so4^-2] = 2.4 x 10^-5 = [0.20][so4^-2] [so4^-2] =1.2 x 10^-4 ag2so4(s) ,> 2ag+ + so4^-2 ksp = [ag+]^2[so4^-2] = 1.5 x 10^-5 = (0.30)^2[so4^-2] [so4^-2] = 1.5 x 10^-5/0.09 = 1.7 x 10^-4 caso4 would precipitate first since it requires less so4^- to form a saturated solution.  i believe this is right, i hope it !  

answer: 88.5%

explanation:

1) write the balanced chemical equation:

2h₂ + o₂ → 2h₂o

2) write the mole ratios:

2 mol h₂ : 1 mol o₂ : 2 moles h₂o

3) calculate the number of moles of each reactant:

formula: number of moles = mass in grams / molar massnumber of moles of h₂: 11.0 g / 2.016 g/mol = 5.46 molesnumber of moles of o₂: 95.0 g / 31.998 g/mol = 2.97 moles

4) determine the limiting reactant:

theoretical ratio: 1 mol o₂ / 2 mol h₂ = 0.5 o₂ / h₂actual ratio: 2.97 mol o₂ / 5.46 mol h₂ ≈ 0.54 o₂ / h₂hence, there is more o₂ moles than the needed to react with the h₂ moles given, so o₂ is in excess and h₂ is the limiting reactant.

5) calculate the theoretical yield using the limiting reactant:

proportion: 2 moles h₂ / 2 moles h₂o = 5.46 moles h₂ / xsolve for x: x = 5.46 moles of h₂oconvert moles to mass of h₂o: 5.46 mol × 18.015 g / mol = 98.4 g

6) calculate the percent yield:

percent yiled = (actual yield / theoretical yield) × 100percent yield = (87.0 g / 98.4 g) × 100 = 88.4% ← answer

note: the 0.1% difference of 88.4% with 88.5% is not relevant since the difference is in the last significant figure and it can be attributed to differences in the precision in the molar masses used.