Chemistry

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x*10^{-4}=6021*10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

$O_2^{-2}=8*2+2=18e^{-}$

${\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2={\pi }_{2py}^2{\pi }_{2px}^{*2}={\pi }_{2py}^{*2}$      

Number of unpaired e^- = 0

Ans. = 0

$\left(\frac{x}{m}\right)=K{P}^{1/n}$           

$\frac{10}{1}=K{\left(100\right)}^{1/n}................\left(1\right)$     

$\frac{15}{1}=K{\left(200\right)}^{1/n}................\left(2\right)$

$\frac{V}{1}=K{\left(300\right)}^{1/n}...............\left(3\right)$

Dividing (2) by (1)

$\frac{15}{10}={\left(\frac{2}{1}\right)}^{1/n}$ = (2)^1/n

$\frac{V}{10}=10^{0.279}$

$V=10*10^{0.279}=10^{1.279}=10^x$

$\therefore x=1.279=127.91*10^{-2}\approx 128*10^{-2}$

Ans. = 128

$C_2{H}_{7}N+\left(2x+\frac{y}{2}\right)CuO\to xC{O}_2+\frac{y}{2}{H}_{2}O+\frac{Z}{2}{N}_2+\left(2x+\frac{y}{2}\right)Cu$

$C_2{H}_7{N}_1+\left(2*2+\frac{7}{2}\right)CuO\to 2C{O}_2+\frac{7}{2}{H}_{2}O+\frac{1}{2}{N}_2+\left(2*2+\frac{7}{2}\right)Cu$              

$\therefore y=7$              

Ans. = 7

Students should prepare complete syllabus when they have time to prepare. However, you can use the list of highweightage chapters in last minute revision for scoring well.

• The p-Block Elements: This chapter holds a high weightage of 8–10 marks. (in the latest syllabus this is deleted)

• Aldehydes, Ketones, and Carboxylic Acids: This chapter contributes around 8–10 marks.

• Biomolecules: This chapter accounts for around 8 marks.

• Chemical Kinetics: This chapter holds a high weightage of 5-6 marks. 

• The d- and f-Block Elements: This chapter contributes around 5-6 marks.

• Amines: This chapter contributes around 5-6 marks.

Kindly consider following image

PV = nRT             PV = constant (at constant T)

Pressure increases & volume decreases, PV remains constant at constant T.

Kindly Consider the following Image

Kindly consider the following Image 

Lactose is polymer contain glycosidic linkage between C₁ of b-galactose & C₄ of b-glucose.