STD 10 SAMAJIK VIGYAN CH 6 SVADHYAY

Calculate the pH of a given mixtures. (a) (4g CH_(3)COOH+6g CH_(3)COONa) in 100 mL of mixture, ( K_(a) for CH_(3)COOH=1.8xx10^(-5)) (b) 5 mL of 0.1MBOH+250 mL of 0.1 MBCI , ( K_(a) for MOH=1.8xx10^(-5)) (c ) ( 0.25 mole of CH_(3)COOH+0.35 mole of CH_(3COOH=3.6xx10^(-4))

How many of the following compounds would give Iodoform test (1) s CH_(3) - CO - CH_(2) - CO - OCH_(3) (2) C_(6) H_(5) - CO - CH_(2) - CO - C_(6)H_(5) (3) CH_(3) - CHCl - CH_(2) - CH_(3) (4) CH_(3) - CHNH_(2) - CH_(3) (5) CH_(3) - CO - CH_(2)I (6) CH_(3) - CO - O - CO - CH_(3) (7) CH_(3) - CO - O - CH_(3) (8) CH_(3) - CO - Cl (9) CH_(3) - CHOH - CH_(2) - CH_(3) (10) CH_(3) - CO - CH_(2) - CHOH - CH_(3)

The reaction C_(6) H_(5) CHO + CH_(3) CHO rarr C_(6) H_(5) CH = CH - CHO is known as

The increasing order of reactivity of the following isomeric halides with Ag NO_(3) (H_(2)O+alcohol) is (I) C_(6) H_(5) -CH=CH - CH_(2) -CH_(2) -CI (II) C_(6) H_(5) -CH =CH-underset(Cl)underset(|)(C)H-CH_(3) (III) C_(6) H_(5) - underset(CI)underset(|)C =CH-CH_(2) -CH_(3) (IV) C_6H_5-underset(Cl)underset(|)overset(C_6H_5)overset(|)C-CH=CH_2

Given the hybridistaiton state of each carbon in the following compounds : I. CH_(2) = C = O II. CH_(3) - CH = CH_(2) III. (CH_(3))_(2) CO IV. CH_(2) = CH = C -= N v. C_(6) H_(6)

The compound that reacts with NaOH . (A) CH_(3)OH (B) CH_(3)CH_(2)OH (C) C_(6)H_(5)OH (D) C_(6)H_(5)CH_(2)OH

Classify the reagents shown in bold in the following equations as nucleophiles or electrophiles : (a) CH_3 COOH + HO^(-) to CH_3 COO^(-) + H_(2) O (b) CH_3 CH_2 Br + HS^(-) to CH_3 CH_2 SH+ Br^(-) (c ) C_6 H_6 + CH_3 CO^(+) to C_6 H_5 COCH_(3) (d) CH_3 COCH_(3) + CN^(-) to CH_3 underset(underset(CN)|)overset(overset(OH)|)C - CH_(3)

Calculate pH of (a) 10^(-1)MCH_(3)COOH (b) 10^(-3)CH_(3)COOH (C )10^(-6)M cH_(3)COOH Take K_(a)=2xx10^(-5) at 25^(@)C .

6 gm of CH_(3)COOH, 6 gm of NaOH and 6.3 gm of HNO_(3) are dissolved in 1L of water. If K_(a) of CH_(3)COOH is 10^(-5) then calculate pH of resulting solution.