Chemistry, 25.11.2020 20:50 cxttiemsp021
Consider this cell: Al(s) | Al3+(aq) || Br2(l) | Br-(aq)1/3Al(s) + 1/2Br2(l) -> 1/3Al3+(Aq) + Br-(aq) Delta H^{0} rxn = -299 kJ/molBr2(l) + 2e - -> 2Br-(aq) : 1.09 VAl3+(aq) + 3e - -> Al(s) : -1.66 V1. What is the standard potential for this cell at 25 degrees C? 2. What is the potential of the cell at room temperature if the concentration of bromide ion is 0.691 M and the concentration of aluminum ions is 31 μM? 3. What is the equilibrium constant for the reaction?4. Without doing any calculations, decide if the standard potential for the cell will increase or decrease if the temperature is increased. Write down your reasoning. What is the standard potential of this cell at 88 °C?
Answers: 3
Chemistry, 22.06.2019 04:30, akeemedwards12
Both josef loschmidt and amedeo avogadro contributed to our understanding of basic molecular numbers, sizes, and reaction ratios. neither scientist discovered “avogadro’s number” in the form we use it today (6.02 x 10 23). still, there’s a controversy over the name. research the contributions from these two scientists and read about how avogadro’s number got its name. briefly state what you think this number should be called, providing key details of each scientist’s contributions to this concept and a solid rationale for your case in naming the number.
Answers: 2
Chemistry, 22.06.2019 20:00, 20calzoy
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. in the first step, calcium carbide and water react to form acetylene and calcium hydroxide: cac2 (s) + 2h2o (g) → c2h2 (g) + caoh2 (s) =δh−414.kj in the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6c2h2 (g) + 3co2 (g) + 4h2o (g) → 5ch2chco2h (g) =δh132.kj calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. round your answer to the nearest kj .
Answers: 3
Consider this cell: Al(s) | Al3+(aq) || Br2(l) | Br-(aq)1/3Al(s) + 1/2Br2(l) -> 1/3Al3+(Aq) + Br-...
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