Interpretation:
The process of zone refining is to be described.
Concept introduction:
The naturally occurring chemical substances in the form of which metals occur in earth, along with impurities, are called minerals.
A mineral from which a metal is economically and conveniently extracted is called an ore. An ore contains many impurities that are removed by purification.
Minerals are the halides, oxides, carbonates, silicates, sulphides, and sulfates of metals.
The impurities of metals are removed by purification processes.
The process of purifying crude metals is called refining.
The purification depends upon the nature of the metal and the nature of the impurities to be removed.
Zone refining is a method to obtain metal with high purity.
Zone refining is based upon the principle that impurities are more soluble in the molten state than in the solid state.
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Chapter 23 Solutions
BURDGE CHEMISTRY VALUE ED (LL)
- A chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forwardNonearrow_forward
- Q2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forward
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