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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

To be eligible for getting a UK work visa, the applicants must have the following:

1. you must have received sponsorship from a licensed sponsor in the UK
2. you must be in the possession of a Certificate of Sponsorship (CoS) with a unique reference number
3. proven English language proficiency by means of getting the required score in standardised tests such as IELTS, TOEFL, PTE and Duolingo, and
4. and enough financial means to support yourself in the UK for the year to come. 

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