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Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$=\left(0.2\times 200\right)\times 10^{-3}-\left(0.1\times 200\right)\times 10^{-3}$                

= 20 × 10^-3 moles

Mass of CH₃COOH absorbed = 20 × 10^-3 × 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6}=2g$ of charcoal.

RRB NTPC preparation plan should be effective and sustaintable. Candidates must prepare different preparation strategy for different stages of recruitment. Given below are few preparation strategy for the exam.

• Understand the structure of the exam
• Know the chapters given in the syllabus 
• Candidates must create a good study plan
• They must take mock tests 
• Practice sample papers that have covered all the chapters given in the syllabus
• Work upon improving the weak areas

$K=\frac{1}{R}\times cellconstant$

= 0.152 × 10^-3 × 1750

 = 266 × 10^-3 cm^-1

Medicaps University offers a comprehensive BSc in Biotechnology course. Students who wish to enrol into the course must have passed Class 12 examinations with Physics, Chemistry and Biology/Maths as compulsory subjects from a recognised board. Students who fulfil this criterion can apply for the course by visiting the official website of the university.

${\left(ta{n}^{-1}x\right)}^3+{\left(co{t}^{-1}x\right)}^3=k{\pi }^3,x\in R$

$\Rightarrow \left(ta{n}^{-1}x+co{t}^{-1}x\right)\left({\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-3ta{n}^{-1}xco{t}^{-1}x\right)=k{\pi }^3$

$\Rightarrow \frac{1}{3}\le k<\frac{7}{8}$

$\begin{array}{l}PC{l}_5\left(g\right)?PC{l}_3\left(g\right)+C{l}_2\left(g\right)\\ t=05moles00\\ \underset{\_}{-xmole}\underset{\_}{+xmole}+\underset{\_}{xmole}\end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$=\left(5-x\right)+x+x+n_{N_2}=\left(7+x\right)$

$K_p=\frac{\left(P_{PC{l}_3}\right)\times {\left(PC{l}_2\right)}_{Cl}}{{\left(P_{PC{l}_5}\right)}_{eq}}=\frac{\left(\frac{3}{10}\times 2.46\right)\times \left(\frac{3}{10}\times 2.46\right)}{\left(\frac{2}{10}\times 2.46\right)}=1.107atm$

= 1107 × 103 atm

Candidates should focus on RRB NTPC preparation for four to six hours daily. This also depends on the current level of preparation and familiarity with the subjects. It is with hardwork and determination that candidates will be able to crack the exam.

There are a lot of similarities between personal and mass communication. Both have message creation, a sender-receiver model, and use various media channels like verbal communication, written text, visual images, audio, and digital platforms. Most importantly, the purpose of both is to establish communication.

The median package offered during VSSUT placements 2025 is yet to be released. Meanwhile, check out the median package offered between 2023 and 2024 in the table below:

Personal communication is divided into various clusters. Some of the four basic categories that come under personal communication are - verbal, listening, written, and nonverbal communication.

According to Henry’s law

$P_{Gas}=K_H\times x_{Gas}$

$=0.835=1.67\times 10^3\times \left[\frac{n_{C{O}_2}}{n_{C{O}_2}+55.5}\right]$

As $n_{C{O}_2}<<55.5$

$\Rightarrow n_{C{O}_2}=27.78\times 10^{-3}moles$

Mass of CO₂ gas = $n_{C{O}_2}\times \left(MM\right)C{O}_2$

= 1222 × 10^-3 gm

$\alpha +\beta =-\frac{\lambda }{3},\alpha \beta =-\frac{1}{3}$

$\therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \lambda =\pm 3$

$\therefore 6{\left({\alpha }^3+{\beta }^3\right)}^2=6{\left({\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right)}^2=6\times {\left(-\left(1+1\right)\right)}^2=24$

Yes, Veer Surendra Sai University of Technology offers decent placements annually for courses like BTech, MTech, MS and MTech. The VSSUT average package offered to the students during the 2025 placement is yet to be released. However, the VSSUT average package offered in 2024 was INR 6.38 LPA. Earlier, the average package offered during VSSUT placements 2023 was INR 6.96 LPA.

Mass communication is the means of transmitting a message or information to the masses. The various examples of mass communication are - radio, television, social networking, billboards, newspapers, magazines, books, film medium, and the internet.

${PV)}_{H_2}={\left(PV\right)}_{Gas}$

${\left(nRT\right)}_{H_2}={\left(nRT\right)}_{Gas}$

$n_{H_2}\times T_{H_2}=n_{Gas}\times T_{Gas}$

$\Rightarrow \frac{0.2}{2}\times 200=\frac{3}{{\left(MM\right)}_{Gas}}\times 300\Rightarrow {\left(MM\right)}_{Gas}=45gm/mole$

Students can start preparing for RRB NTPC exam by going through the exam pattern. It also necessary to know syllabus. Candidates must also follow good study material. They should practice question papers.

The various types of mass communication are radio, television, newspapers, magazines, internet, while types of personal communication are - memos, to-do lists, reminders, drafts, one-on-one interviews, and telephonic conversations.

Communication is the way to share and receive messages and information between two or more people. A successful communication needs a sender, a message, and a receiver.

Communication are of two types - mass communication, and personal communication. These are then divided into two parts verbal and nonverbal communication and visual communication.