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Yes, joining the best Perfusion Technology colleges in India can be worthwhile for you. For that, you need to check its ROI. Listed below are some colleges along with their tuition fees and average placement package:

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Chlorine nitrate as hydrolysis gives hypohalous acid and nitric acid as,

$C\mathcal{l}ON{O}_2+H_{2}O\to \underset{A}{HOC\mathcal{l}}+\underset{B}{HN{O}_3}$                

Chlorine nitrate on reaction with HCl produces Cl₂ and HNO₃ as,

$C\mathcal{l}ON{O}_2+HC\mathcal{l}\to \underset{C}{C{\mathcal{l}}_2}+\underset{B}{HN{O}_3}$            

Yes, there are about 8 top Perfusion Technology colleges in India that offer Diploma courses to their students:

Disclaimer: This information is sourced from official website and may vary.

Students at Sheridan can choose from more than 140+ options if they're looking for a UG or PG degree. Apart from Business Management, IT, and Software Engineering, some other popular courses at the institute include:

• Design
• Law
• Mass Communication and Media
• Humanities
• Arts
• Science
• Architecture
• Accounting
• Medical Science
• Education

Yes, there are many affordable Perfusion Technology colleges in India. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from official website and may vary.

As per popularity basis, listed below are the top Perfusion Technology colleges in India along with their tuition fees:

Disclaimer: This information is sourced from official website and may vary.

About 80+ best Perfusion Technology colleges are there in India. Some of the top Perfusion Technology colleges in India include Postgraduate Institute of Medical Education and Research, JIPMER Puducherry, AIIMS Rishikesh, Sanjay Gandhi Postgraduate Institute of Medical Sciences, All India Institute of Medical Sciences, Bhubaneswar, Sree Chitra Tirunal Institute for Medical Sciences and Technology, etc.

Of these, 57 colleges are privately owned and 17 colleges are owned by public/government organisations. Students can secure admission on merit basis or through entrance exam like KCET, CUET, etc.

Electronic configuration of the given ions is :

$E{u}^{2+}-4f^{7}6s^0$                 

$\begin{array}{l}T{m}^{2+}-4f^{13}6s^0\\ S{m}^{2+}-4f^{6}6s^0\\ T{m}^{3+}-4f^{12}6s^0\end{array}$

$\begin{array}{l}T{b}^{4+}-4f^{7}6s^0\\ Y{b}^{2+}-4f^{14}6s^0\\ D{y}^{3+}-4f^{9}6s^0\\ y{b}^{3+}-4f^{13}6s^0\end{array}$              

Hence, Eu²⁺ and Tb⁴⁺ have half | filled f-orbitals and Yb²⁺ have completely filled f-orbitals.

In 3d-series, all metals except Cu have negative value of $E_{M^{2+}/M}^0$ due to low hydration enthalpy and high I.E and sublimation enthalpy.

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

$N{H}_{4}C\mathcal{l}\left(aq\right)+NaN{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(\mathcal{l}\right)+NaC\mathcal{l}\left(aq\right)$  

Hence, N₂ gas is produced.

For isoelectronic species

$Ionicradii\propto \frac{Numberofe}{Numberofp}$              

$\begin{array}{l}N{a}^{+}M{g}^{2+}{N}^{3-}{O}^{2-}{F}^{-}\\ e=1010101010\\ p=1112789\end{array}$                

Hence, correct order of ionic radii is;

$M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$                

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH₄ < HCN < NH₃

Leaching involves the given reaction,

$4Au\left(s\right)+8C{N}^{-}\left(aq\right)+2H_{2}O\left(aq\right)+O_2\left(g\right)\to 4{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+4O{H}^{-}\left(aq\right)$              

Here, O₂ is required for formation of Au (l) cyanide complex but no complex in absence of O₂.

$2{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+Zn\left(s\right)\to {\left[Zn{\left(CN\right)}_4\right]}^{2-}\left(aq\right)+2Au\left(s\right)$                

In above displacement reaction, Zn is oxidized.

Structure PCl₅ is trigonal bipyramidal,

Hybridization of P is sp³d

Equatorial bonds lie in a plane

Axial bonds are longer than equatorial bonds so axial bonds are weaker than equatorial bonds.

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH₄ < HCN < NH₃