Class 12th

Density, $\rho =7.62gc{m}^{-3}$  

Cell edge, a = 0.4518nm = 0.458 * 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $\rho =\frac{Z*A}{N_A*a^3}$  

$\rho =\frac{4*A}{6.022*10^{23}*{\left(0.458*10^{-7}\right)}^3}$              

$7.62=\frac{4*A}{6.022*{\left(0.4518\right)}^3*10^2}$              

A = 105.79 g/mol

Ans. is 106 (the nearest integers)

It is given that they were paid in the ratio 4 : 5 : 6 s for doing the job.

So, maximum amount would have been got by z and the minimum amount obtained by x.

Amount obtained by x = $\frac{}{}$ $=4*\frac{T}{15}$

where T is the total amount that all of them got together.

Amount obtained by z   $=6*\frac{T}{15}$ $\frac{}{}$

Now as per the problem,

$=6*\frac{T}{15}-4*\frac{T}{15}=300$

$=2*\frac{T}{15}=300$

T = Rs. 2250

Share of x and z $=6*\frac{T}{15}+4*\frac{T}{15}=10*\frac{T}{15}=$ $\frac{}{}\frac{}{}\frac{}{}$ Rs. 1500

2¹⁸⁹ – 2¹⁸⁸ – 2¹⁸⁸ = 2¹⁸⁸ (2² – 2 – 2) = 0

$\frac{100}{3}\mathrm{\%}\text{?}\text{?}=\text{?}\text{?}\frac{x}{1000\text{?}\text{?}?\text{?}\text{?}x}\text{?}\text{?}*\text{?}\text{?}100\mathrm{\%}$

x = 250

Hence, he uses weight, 1000 – 250 = 750 g

For first order reaction, using

$K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$              

$K=\frac{2.303}{60}\left(0.477-0.301\right)mi{n}^{-1}$              

$K=6.7*10^{-3}mi{n}^{-1}$              

Ans. is 7 (the nearest integer)

Let other number be x.

6 * 84 = 12 * x

x = 42

the highest power =   $\frac{80}{9}$  = 8

NH₃ behaves as strong field ligand with Co²⁺ & Co³⁺

$\left[Co{\left(N{H}_3\right)}_6\right]C{l}_2$              

$C{o}^{2+}-4s^{0}3d^7$      

It has one unpaired electorn.

 $\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$

$C{o}^{3+}-4s^{0}3d^6$    

It has no unpaired electrons

So, total unpaired electrons in both compounds is 1.

HCF = $\frac{HCFof(81,27,9)}{LCMof(16,20,40)}=\frac{9}{80}$  

Quantitative Aptitude Prep Tips for MBA

$\begin{array}{l}=\sqrt[]{(l^2+b^2+h^2)}\\ =\sqrt[]{100+100+225}\\ =\sqrt[]{425}=5\sqrt[]{17} m\end{array}$