Chemistry

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V

W = 4.75 g

$n=\frac{4.75}{26}$ T = 323 K, R = 0.0826

  $P=\frac{740}{760}atm$          

 V = $\frac{nRT}{P}=\frac{4.75*0.0826*323}{26\left(\frac{740}{760}\right)}=5litre$

$\frac{1.86}{93}=\frac{w}{135}$

W = 2.70 g

Mass produced actual = 2.70 $*\frac{90}{100}=2.43=243*10^{-2}g$

$C_x{H}_y\left(g\right)+\left(x+\frac{y}{4}\right)O_2\left(g\right)\to xC{O}_2\left(g\right)+\frac{y}{2}{H}_{2}O\left(\mathcal{l}\right)$

V 6V - -

- - 4V

Volume of CO2 = 4 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

Vx = 4V

x = 4

Volume of O2 = 6 * $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

$V\left(x+\frac{y}{4}\right)=6V$

$4+\frac{y}{4}=6$

y = 8

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0*10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8*10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8*10^{-9}=0.1*{\left(2S\right)}^2$

S = 141 * 10^-6 M

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

1° Amine

CH₃ – CH₂ – NH₂

Kindly consider the following figure

Kindly consider the following figure

Linear (180°)