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$Mn{O}_4^{-}+S_2{O}_3^{2-}\stackrel{O{H}^{-}}{\to }Mn{O}_2+S{O}_4^{2-}$

O.S of S in    $S{O}_4^{2-}=+6$

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

Ratio = n/n = 1

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• a UK Home Office approved employer must sponsor your work visa and grant you a Certificate of Sponsorship  
•  you must prove proficiency in English language - given that you have not been living in the country for last 1 year.  Recent graduates or holders of UK PSW Visa do not need to meet this requirement  

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$         

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$  

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6\times \Delta {H^0}_f\left(F\right)$            

= (-1100) - (275) -6 (80) = -1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$            

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii) $a^2-b^2=48$  

a² = 75, b² = 27

Seliwanoff's test is used to distinguish carbohydrates while xanthoproteic test is used to distinguish proteins

If candidates have focused well in their 10 and 10+2 classes, they can easily qualify in the exam. The question paper is similar to popular engineering entrance exams such as JEE Mains, but the difficulty level is of easy to moderate.

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)           

6x = 5 = 0             $x=\frac{5}{6}$  

P (atmost 2H)

$P\left(OH,5T\right)+P\left(1H,4T\right)+P\left(2H,3T\right)$  

$=\frac{1}{6^5}\left(1+25+250\right)=\frac{276}{6^5}=\frac{4^6}{6^4}$  

sin x = 1 – sin² x

=> sin x = $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y = $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}\times \left(\overrightarrow{c}\times \overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}\times \left(\overrightarrow{b}\times \overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

First common term to both AP’s is 9

t₇₈ of $\left(3,6,9,......\right)=78\times 3=234$  

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$  

n^th common term $\le 234$  

9 + (n – 1) 12 $\le$  234

n < $\frac{237}{12}\Rightarrow n=19$  

Now sum of 19 terms with a = 9, d = 12

  $=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$

Species               B.O

Ne₂                        0

N₂                           3

F₂                 1

O₂                          2

At the TS EAMCET exam centre, candidates need to follow the below guidelines - 

• Reach the exam centre atleast 40-60 minutes before the reporting time.
• Carry your government issued IDs.
• Carry a passport size photo,
• Although the exam is in CBT based online mode, carry black and blue ball point pen.
• Avoid wearing flashy things such as jewellery at the exam centre.
• Avoid taking your smartphone, smartwatch to the exam centre.

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y = $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$

y(2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$ $\frac{{}^{}}{{}^{}}\frac{}{}$

y(3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

Data contradiction.

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

Kindly consider the following figure

2, 4-DNP (Brady's Reagent) is used to test carbonyl compound