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Shahdara, Delhi Details verified of Sagardeep Singh✕
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Share this profile Hindi Proficient
English Proficient
Punjabi Basic
Indraprastha University 2018
Bachelor of Technology (B.Tech.)
Shahdara, Delhi, India - 110032
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
3
Board
CBSE
Subjects taught
Science, English, Mathematics, Social Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
3
Board
CBSE
Subjects taught
English, Mathematics, Science, Social Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
4
Board
CBSE
Subjects taught
Mathematics, English, Computers, Science, EVS
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 7 Tuition
4
Board
CBSE
Subjects taught
Science, Mathematics, EVS, English, Computers
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
4
Board
CBSE
Subjects taught
Computers, Science, EVS, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
CBSE
Subjects taught
English, Mathematics, Physics, Economics, Chemistry
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
CBSE
Subjects taught
Chemistry, Economics, Mathematics, Physics, English
Taught in School or College
No
Answered on 08/11/2018 Learn CBSE - Class 11/Chemistry
Ask a Question
Post a Lesson
It's a simple solution. Just take molar mass of isotopes, mulitply by the percentage and you get your answer.
{35.96755 × (0.337/100)} + {37.96272 × (0.063/100)} + {39.9624 × (99.600/100)} =
0.1212106435 + 0.0239165136 +39.8025504 =
39.947677557
Like 0 
Answers 1 
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
3
Board
CBSE
Subjects taught
Science, English, Mathematics, Social Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
3
Board
CBSE
Subjects taught
English, Mathematics, Science, Social Science
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
4
Board
CBSE
Subjects taught
Mathematics, English, Computers, Science, EVS
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 7 Tuition
4
Board
CBSE
Subjects taught
Science, Mathematics, EVS, English, Computers
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
4
Board
CBSE
Subjects taught
Computers, Science, EVS, English, Mathematics
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
2
Board
CBSE
Subjects taught
English, Mathematics, Physics, Economics, Chemistry
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
2
Board
CBSE
Subjects taught
Chemistry, Economics, Mathematics, Physics, English
Taught in School or College
No
Answered on 08/11/2018 Learn CBSE - Class 11/Chemistry
Ask a Question
Post a Lesson
It's a simple solution. Just take molar mass of isotopes, mulitply by the percentage and you get your answer.
{35.96755 × (0.337/100)} + {37.96272 × (0.063/100)} + {39.9624 × (99.600/100)} =
0.1212106435 + 0.0239165136 +39.8025504 =
39.947677557
Like 0 Answers 1 Comments Ask a Question
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