write the lewis structures of the reactants and pr...
write the lewis structures of the reactants and product of each of the following equations, and identify the lewis acid and the lewis base in each:(a) cs2 + sh− ⟶ hcs3−
we’re being asked which of the given compounds is a lewis acid and lewis base. let’s first define a lewis acid and a lewis base.based on the lewis definition:lewis acid is an electron pair acceptor.some characteristics of a lewis acid:• when hydrogen is connected to an electronegative element such as p, o, n, s or halogens▪ hydrogen gains a partially positive charge → makes hydrogen act as a lewis acid• positively charged metal ions[readmore]▪ can accept negative electron pairs → can act as lewis acids• central element that has less than 8 valence electrons around it▪ more likely to accept an electron pair to follow the octet rule▪ can accept electron pair → can act as lewis acidlewis base is an electron pair donor.some characteristics of a lewis base:• compounds that have lone pair(s) on a central element can act as lewis bases. • compounds that possess a negative charge can act as lewis bases. let’s draw the lewis structure of each given compound and check which of them can act as a lewis acid and lewis base.step 1: determine the central atom in this molecule.step 2: calculate the total number of valence electrons present.   step 3: draw the lewis structure for the molecule.a) cs2step 1: determine the central atom in this molecule.c is less electronegative than s → c will be the central atomstep 2: calculate the total number of valence electrons present.                group             valence electronsc         4a                   1 x 4 e‑ = 4 e-s         6a                   2 x 6 e‑ = 12 e-              total: 16 valence e- step 3: draw the lewis structure for the molecule• c has a total of 8 e- electrons around it and has a lone pair→ complete octet▪ can act as a lewis acidb) sh−step 1: determine the central atom in this molecule.only two atom → no central atomstep 2: calculate the total number of valence electrons present.                group             valence electronss         6a                   1 x 6 e‑ = 6 e-h        1a                   1 x 1 e‑ = 1 e-from -1 charge:                     +1 e-              total: 8 valence e- step 3: draw the lewis structure for the molecule▪ has a negative charge▪ can act as a lewis baseanswer: sh- is the lewis base and its lewis structure is cs2 is the lewis acid and its lewis structure is
a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 consider lhs,lhs = secθ (1 – sinθ) (secθ + tanθ)⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ - ) (secθ + tanθ)we know that = tanθ.⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ – tanθ) (secθ + tanθ)we know that (a + b) (a – b) = a2 – b2.⇒ secθ (1 – sinθ) (secθ + tanθ) = sec2θ – tan2θwe know that sec2θ – tan2θ = 1.∴ secθ (1 – sinθ) (secθ + tanθ) = 1 = rhshence proved. dear student

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regards write the reactions of f2 with h2o? hint: at first think about the structure of ester so that you are able to draw the structure for a given molecular formula. the structure of ester is $r - co - o - r'$.complete answer:an ester is a chemical compound derived from an acid in which at least one –oh group is replaced by an $ - o - $ alkyl group. esters are generally derived from substitution reactions of a carboxylic acid and an alcohol which is also known as an esterification reaction. from the given molecular formula we can draw only two structures of esters. they are ethyl formate and methyl acetate. their structures are $h - co - o - c{h\_2} - c{h\_3}$ and $c{h\_3} - co - o - c{h\_3}$.additional information:esterification is the process of combining carboxylic acid and alcohol to give at least one ester and water.$r - co - oh + r - oh \rightleftharpoons r - co - o - r' + {h\_2}o$in the above reaction, the forward reaction gives us esterification and the backward reaction gives us hydrolysis of the ester. the hydrolysis of esters in the presence of alkalis such as potassium hydroxide (lye) or sodium, a reaction called saponification which is utilized in the preparation of soaps from fats and oils and is also used for the quantitative estimation of esters.note:esters are common in organic chemistry and biological materials, and often have a pleasant characteristic, fruity odor. this leads to their extensive use in the fragrance and flavor industry. ester bonds are also found in many polymers. given, x3 – 1, x – 1 as pair of polynomialsneed to find the quotient and remainder⇒ to find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b⇒ since, the polynomial is of third degree we can write the q(x) as x2 + ax + b∴ p(x) = (x – a)( x2 + ax + b) + c⇒ x3 + 1 = (x + 1)(x2 + ax + b) + c⇒ x3 + 1 = (x3 + x2 + ax2 + ax + bx + b) + c⇒ x3 + 1 = x3 + (a + 1)x2 + (b + a)x + (c + b)∴ a + 1 = 0, b + a = 0, c + b = 1⇒ a = – 1, b = 1⇒ c + b = 1⇒ c = 0quotient = x2 + ax + b = x2 – x + 1remainder = 0 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 nucletic acid `overset("nucelase")rarr` nucleotides `overset"nuceotidase")rarr`nucleosides `overset("nucleosidase")rarr` sugars+bases 
soucrose `overset("sucrase")rarr` glucose + fructose 
fats `overset("lipase")rarr` diglycerides `overset("lipase")rarr` monoglycerides 
proteins, peptones, proteoses `underset("carboxypeptidase")overset("trypsin//chymotrypsin")rarr` dipeptides
answer: c
solution: bile salts of bile bring about emulsification of fat by which fat droplets are broken down into may small droplets this increaes lipase action on fat, by increasing the available surface area. solid + liquid mixture : sand and water, charcoal and water. use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4