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**K**< Cs1 Answers · Science & Mathematics · 29/09/2020

q = Ccal (T2-T1) 2.24 kJ = (0.827 kJ/ C ) (T2 - 23.44° C ) T2 = 26.15 ° C

1 Answers · Science & Mathematics · 01/10/2020

... × (V₂/V₁) = (273 + 40) × (6.50/2.30) K = 885 K = 612° C ==== Hydrogen gas was cooled from 150° C to 50° C ...

1 Answers · Science & Mathematics · 21/07/2020

a) 2 K {+} + CO3{2-} + Pb{2+} + 2 NO3{-} → PbCO3(s) + 2 K {+} + 2 NO3{-} b) CO3{2-} + Pb{2+} → PbCO3(s) c ) K {+} and NO3{-}

1 Answers · Science & Mathematics · 11/06/2020

see the pic where I worked out the conversions for you. both A and B are correct answers.

4 Answers · Science & Mathematics · 25/10/2020

... is 1 atm, V1 is 1.00 L, T1 is 273

**K**, P2 is 2.00 atm, V2 is 10.0 L, solve for ...2 Answers · Science & Mathematics · 14/08/2020

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**c**is being produced = 1*1.2 M/s = what? then .. rate =**k*** [0.15M] = rate A = rate**C**= what?1 Answers · Science & Mathematics · 16/07/2020

V = nRT / P = (12.3 g Cl2 / (70.9064 g Cl2/mol)) x (8.3144626 L kPa/

**K**mol) x (13 + 273)**K**/ (96.4 kPa) = 4.28 L1 Answers · Science & Mathematics · 28/05/2020

ΔrG° = - RT ln Keq ΔrG° = - 8.314 J/molK (273.15 + 31.1 K ) ln (2.3X10^3) = 1.96X10^4 J/mol = 19.6 kJ/mol

1 Answers · Science & Mathematics · 11/04/2020

ΔG° = -RT ln Keq ΔG° = - 8.314 J/molK (329.25K) ln (2.9X10^8) = -5.33X10^4 J/mol = -53.3 kJ/mol

1 Answers · Science & Mathematics · 07/08/2020