Answer to Question #49762 in Inorganic Chemistry for pink




 how do you write electron configurations in short hand form
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we are asked to write the chemical formula of hydrates and to use proper symbols, case, and formatting.remember that an ionic hydrate is formed when an ionic compound is bonded to one or more molecules of water. naming ionic hydrates has the same set of rules in naming the ionic compound portion. the water molecule portion is named with a numerical prefix together with the word hydrate.[readmore]step 1: discuss the rules in naming ionic hydrates.step 2: from the naming rules, extract the chemical symbols and the formula for each of the compounds.step 1: discuss the rules in naming ionic hydrates.rules in naming the ionic compound portion:1.a. the metal is named and normally written first.-if the metal is a transition metal we must use a roman numeral to describe its positive charge.1.b. the nonmetal keeps the first part of its name but has its ending changed to -ide.1.c. name the polyatomic as you would normally.rules in naming the h2o portion:1.d. the h2o portion will be called hydrate.1.e. to describe the number of h2o molecules, use the numerical prefixes.note: a dot (•) is written in between the ionic compound and the hydrate.step 2: from the naming rules, extract the chemical symbols and the formula for each of the compounds.a. iridium(iii) bromide tetrahydrate rule 1.a.iridium→transition element → symbol : iriii → describes the transition element's positive charge → symbol: 3+rule 1.b.bromide → nonmetal bromine → bromide ion has a charge of -1 → symbol: brwe write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.we don't need to write 1 as a subscript, it is understood to be 1 without it.rule 1.d.hydrate → h2orule 1.e.tetra → four → 4therefore the chemical formula is irbr3•4h2o.b. tin(iv) chloride pentahydraterule 1.a.tin(iv)→ sn4+rule 1.b.chloride → cl-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.again, we don't need to write 1 as a subscript.rules 1.d and 1.e.pentahydrate → 5 h2o the chemical formula is sncl4•5h2o.therefore the chemical formula is sncl4•5h2o.c. zinc nitrate hexahydrate rule 1.a.zinc→ zn1+rule 1.c. nitrate → no31-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.note that +1 and -1 charges just cancel out, we just put the atoms together.rules 1.d and 1.e.hexahydrate → 6 h2othe chemical formula is znno3•6h2o.therefore the chemical formula is znno3•6h2o.d. sodium carbonate monohydraterule 1.a.sodium→ na1+rule 1.c. carbonate → co32-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.again, we don't need to write 1 as a subscript.rules 1.d and 1.e.monohydrate → 1 h2o → h2o     (we don't need to write 1 as a coefficient.)the chemical formula is na2co3•h2o.therefore the chemical formula is na2co3•h2o. chapter 5 key: electron: e                         number: # ­                 valence electron: ve   three­dimensional: 3d molecular: molec               with: w/ chemical: chem                 different: diff positive: pos negative: neg i. bonding theories ­ a. valence shell e pair repulsion (vsepr)  valence shell e pair repulsion: model predicting the arrangements of ve  ­ pairs around a central atom that minimizes their mutual repulsion  electron pair geometry: 3d arrangements of bonding pairs andn lone pairs  of e about a central atom  molecular geometry: 3d arrang write the reactions of f2 with h2o? <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> according to ncert answer would be option (1)and

but according to general chemistry answer would be (3)andoption 1)and





option 2)andoption 3)andoption 4)and cao the q value for the nernst equation can be determined by  the overall reaction of the cell notation. recall that q is the reaction quotient which can be calculated as:determine the overall reaction using the mnemonic for cell notation, abc:step 1. determine the overall reaction and find the anode and cathode. the anode (one being oxidized) and cathode (one being reduced). using the mnemonic, it appears that the anode is zn(s) while the cathode is cu(s).step 2. write the equation where the anode, zn (s), is being oxidized to to zn2+ (from zn(no3)2) and the cathode, cu (s), is produced by reduction of cu2+ (from cu(no3)2zn (s) - anode; cu (s) - cathode zn (s) + cu2+ (aq) → cu (s) + zn2+ (aq)step 3. setup the q expression using the concentration of zn2+ and cu2+ provided. note that (s) and (l) are not included in q or keq expressionsstep 4. plugin q to lnq ln(q) = ln(1) = 0 answer is 0 there are 3 types of hybridisation for carbon in organic compounds, sp, sp2 and sp3 depending on the number of sigma and pi bonds between the carbon atoms. more the ‘s’ character more is the electronegativity of the carbon as ‘s’ orbitals are closer to the nucleus, hence electrons held more closely. direction: a word and number arrangement machine when given an input line of word and numbers rearranges them following a particular rule in each step. the following is an illustration of input and rearrangement. input: politics 61 policy 116 partner 93 role 45 clear 27 38 point step i: role politics policy 116 partner 93 45 clear 27 38 point 61 step ii: role clear politics policy partner 93 45 27 38 point 116 61 step iii: role clear partner politics policy 93 45 38 point 27 116 61 step iv: role clear partner politics policy 93 38 point 45 27 116 61 step v: role clear partner politics point policy 93 38 45 27 116 61 step v, is the last step of the above arrangement as the intended arrangement is obtained. as per the rule followed in the given steps find out the appropriate steps for the given input. input: energy 54 78 place 23 issue 211 145 flat rent taxi 17in step iii, what is the difference between fifth element from left and second element from right? we are going to do the following steps:step 1: determine the classification of the compoundstep 2: write the formula of the compound[readmore]step 1: determine the classification of the compoundrecall the definition of the following:ionic compound - usually made up of a metal and a non-metalcovalent compound - composed of more than one nonmetal togetheracid - substance whose molecules yield hydrogen ions (h+) when dissolved in waterhydrate - substance that contains water or its constituent elementsfor cobaltic iodidecobaltic – metaliodide – nonmetalmetal + nonmetal = ionicstep 2: write the formula of the compoundthe rules for naming ionic compounds are as follows:1. the metal keeps its name and is named and written first.– for transition metals, we use roman numerals to denote its charge.2. the nonmetal keeps its base name but has its ending changed to –ide.for cobaltic iodidecobaltic  → transition metal → +2 charge → co2+iodide → group 7a → -1 charge → i-criss-cross the charges and make them the subscript of the other ioncobaltic iodide → coi2answer: cobaltic iodide is an ionic compound with a formula of coi2 this is not perfect but i hope it shows you one possible approach. i hand calculated the position of the whisker of each box plot and use that to place the y position of the labels.

plot<-q + scale\_x\_discrete(breaks=c("b","h"), labels=c("bs", "ah"))

plot2<-plot + theme(axis.title.x = element\_blank(…