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For example, sulfuric acid (H 2 SO 4) is a diprotic acid. Since only 0.5 mol of H 2 SO 4 are needed to neutralize 1 mol of OH −, the equivalence factor is: f eq (H 2 SO 4) = 0.5. If the concentration of a sulfuric acid solution is c(H 2 SO 4) = 1 mol/L, then its normality is 2 N. It can also be called a "2 normal" solution.
where A and B are reactants C is a product a, b, and c are stoichiometric coefficients,. the reaction rate is often found to have the form: = [] [] Here is the reaction rate constant that depends on temperature, and [A] and [B] are the molar concentrations of substances A and B in moles per unit volume of solution, assuming the reaction is taking place throughout the volume of the ...
The 2014 AP Chemistry exam was the first administration of a redesigned test as a result of a redesigning of the AP Chemistry course. The exam format is now different from the previous years, with 60 multiple choice questions (now with only four answer choices per question), 3 long free response questions, and 4 short free response questions.
The solution has 1 mole or 1 equiv Na +, 1 mole or 2 equiv Ca 2+, and 3 mole or 3 equiv Cl −. An earlier definition, used especially for chemical elements , holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen , 8 g (0.28 oz) of oxygen , or 35.5 g (1.25 oz) of chlorine —or that will displace ...
The exam is composed of 2 sections, each with 2 different types of questions. Section I consists of 40 multiple choice questions. 28 do not allow the use of a calculator, while the last 12 do allow a calculator. The non-calculator section is worth 43.75% of the exam score, while the calculator section is worth 18.75%. [5]
5 Egg Myths Dispelled By An Expert, Plus Tips For Every Egg Lover. But if the membrane is broken, it is not safe for humans to eat, said Steele. It is, however, still good for the compost bin.
The Pt anode generates I 2 from the KI when current is provided through the electric circuit. The net reaction as shown below is oxidation of SO 2 by I 2. One mole of I 2 is consumed for each mole of H 2 O. In other words, 2 moles of electrons are consumed per mole of water. 2 I − → I 2 + 2 e − B·I 2 + B·SO 2 + B + H 2 O → 2 BH + I ...
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