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Disproportionation reaction is

$\stackrel{+6}{Mg}{O}_4^{--}+4H^{+}\to 2\stackrel{+7}{Mn}{O}_4^{-}+\stackrel{+4}{Mn}{O}_2+2H_{2}O$

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The  SSC JE cut-off is the minimum score that a candidate must obtain in order to qualify for the next stage of the selection process or for final selection. The cut-off is decided by the commission based on various factors such as the number of candidates appearing for the exam, the difficulty level of the exam, and the number of vacancies available for the post. The cut-off may also vary for different categories and posts.

Regarding Lassaigne's Test, both statement are correct.

Kindly consider the following figure

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Be does not perform in flame colouration due to highly ionization energy.

The SSC JE 2025 application process started on June 30, and the last date to apply was July 21, 2025. Candidates can apply by visiting the official website at ssc.gov.in. 

During removal of temporary hardness of water.

$\begin{array}{l}Mg{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }Mg{\left(OH\right)}_2+2C{O}_2\uparrow \\ Ca{\left(HC{O}_3\right)}_2\stackrel{Boil}{\to }CaC{O}_3+H_{2}O+C{O}_2\end{array}$

In sulphide ore, depressants selectively prevent impurity from coming to the fourth.

Yes. The authority released the notification and application form. The last date to apply was July 21, 2025. The SSC JE Tier 1 exam 2025 was conducted from December 3 to 6. 

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Except Te, all are metals.

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^2}{v}_1$

$\Rightarrow \left(\frac{L_2}{T_2}\right)=\frac{n}{m_2}\left(\frac{4}{T_1}\right)$ $\left[?v_2=\frac{L_2}{T_2},v_1=\frac{L_1}{T_1}\right]$

$\Rightarrow \frac{L_2}{L_1}=\left(\frac{n}{m_2}\right)\left[\frac{T_2}{T_1}\right]--\left(i\right)$

And a2 = $\frac{a_1}{mn}\Rightarrow \frac{v_2}{T_2}=\frac{v_1}{T_{1}mn}$

$\Rightarrow \frac{T_1}{T_2}=\frac{v_1}{v_2}\left(\frac{1}{mn}\right)$ $\left[?\frac{v_1}{v_2}=\frac{m^2}{n}\right]$

$\Rightarrow \frac{n^2}{m}{T}_1=T_2$

from (1)

$\frac{L^2}{L_1}=\frac{n}{m^2}\left[\frac{n^2}{m}\right]\left[?\frac{T_2}{T_1}=\frac{n^2}{m}\right]$

$\Rightarrow \frac{n^3}{m^3}{L}_1=L_2$

Invertase Cane sugar to glucose and fructose

Zymase Glucose to ethanol

Diastase Starch to Maltose

Maltase Maltose to glucose

HOCl produce in the stratospheric cloud, by the hydrolysis reaction of ClONO₂.

$ClON{O}_{2\left(g\right)}+H_2{O}_{\left(g\right)}\to HOC{l}_{\left(g\right)}+HN{O}_{3\left(vap\right)}$

The mode of the Institute of Engineering and Management course admissions is online. IEM Kolkata admission to various UG and PG courses is based on entrance examination scores and merit. The accepted entrance exams are WBJEE,  JEE Main, CAT, and others. Candidates willing to get IEM Kolkata admission must take part in the IEM College Kolkata admission process by applying online. First, students must visit the official website. Then, click on the Apply Now tab. Then, select the required course. Then, pay the application fee. Then, fill out and submit the application form with the required information. 

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$

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