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Shweta S.

Dhantoli, Nagpur, India - 440010

Shweta S. Class 6 Tuition trainer in Nagpur

Shweta S.

M. Sc in physics , teaching since 2 years at home

Dhantoli, Nagpur, India - 440010.

Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

I am a student who has passed her m.sc from rtmnu in 2018. I have a basic understanding of concepts of physics of 12th and B.Sc which forms a foundation in many further postgraduate degrees and courses I can teach these concepts in an easy and flexible way so that my students can grasp concepts rather than just rote- reading.

Languages Spoken

English Proficient

Hindi Proficient

Education

RASHTRASANT TUKADOJI MAHARAJ NAGPUR UNIVERSITY 2018

Master of Science (M.Sc.)

Address

Dhantoli, Nagpur, India - 440010

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Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

Reviews

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FAQs

1. Which school boards of Class 8 do you teach for?

CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach BSc Tuition, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition and Class I-V Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for less than a year.

Answers by Shweta S. (1)

Answered on 28/11/2018 CBSE/Class 11/Science/Chemistry/Equilibrium CBSE/Class 11/Science/Chemistry

The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic... ...more
The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.

disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+) CH3COOH ⇔CH3COO- + H+ given: disassociation or ionization constant , K=1.74 ×10 -5 concentration, C =0.05 M So,degree of disassociation or concentration of acetate ions (α) is K=α2C ⇒α... ...more

disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+)

CH3COOH ⇔CH3COO+ H+

given: disassociation or ionization constant , K=1.74 ×10 -5

concentration, C =0.05 M

So,degree of disassociation or concentration of acetate ions (α) is

K=α2C ⇒α =√(K/C)

α =√(1.74 ×10-5/0.05) =√ 0.000348=0.01865

also ,[H+] =αC =0.01865 ×0.05=9.325×10-4

So pH = -log10[H+]= -log10(9.325×10-4)=3.03

Hence ,pH of solution is 3.03

Answers 1 Comments
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Class 6 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

Class 7 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

Class 8 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

Class 9 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

Taught in School or College

No

State Syllabus Subjects taught

Science, English, Mathematics

Class 10 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Science, Mathematics, English, Hindi

Taught in School or College

No

State Syllabus Subjects taught

Science, English, Hindi, Mathematics

Class 11 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

Taught in School or College

No

State Syllabus Subjects taught

English, Physics, Mathematics, Chemistry, Biology

BSc Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

BSc Physics Subjects

Electromagnetic Theory, Solid State Physics

Type of class

Regular Classes, Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

BSc Branch

BSc Physics

Class 12 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

State, CBSE

CBSE Subjects taught

Physics, Chemistry, English, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

English, Physics, Mathematics, Chemistry, Biology

Class I-V Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State

CBSE Subjects taught

Science, English, Hindi, EVS, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics, English, Hindi, Science, Social Science, EVS

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Answers by Shweta S. (1)

Answered on 28/11/2018 CBSE/Class 11/Science/Chemistry/Equilibrium CBSE/Class 11/Science/Chemistry

The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic... ...more
The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.

disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+) CH3COOH ⇔CH3COO- + H+ given: disassociation or ionization constant , K=1.74 ×10 -5 concentration, C =0.05 M So,degree of disassociation or concentration of acetate ions (α) is K=α2C ⇒α... ...more

disassociation of acetic acid takes place as acetate ions (CH3COO-) and hydrogen ions (H+)

CH3COOH ⇔CH3COO+ H+

given: disassociation or ionization constant , K=1.74 ×10 -5

concentration, C =0.05 M

So,degree of disassociation or concentration of acetate ions (α) is

K=α2C ⇒α =√(K/C)

α =√(1.74 ×10-5/0.05) =√ 0.000348=0.01865

also ,[H+] =αC =0.01865 ×0.05=9.325×10-4

So pH = -log10[H+]= -log10(9.325×10-4)=3.03

Hence ,pH of solution is 3.03

Answers 1 Comments
Dislike Bookmark

Shweta S. describes herself as M. Sc in physics , teaching since 2 years at home. She conducts classes in BSc Tuition, Class 10 Tuition and Class 11 Tuition. Shweta is located in Dhantoli, Nagpur. Shweta takes Regular Classes- at her Home. Shweta has completed Master of Science (M.Sc.) from RASHTRASANT TUKADOJI MAHARAJ NAGPUR UNIVERSITY in 2018. She is well versed in English and Hindi.

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