Dilsukhnagar, Hyderabad Details verified of Ashish✕
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Pondicherry University Pursuing
Bachelor of Science (B.Sc.)
Dilsukhnagar, Hyderabad, India - 500059
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Social Science, Hindi, English, Science, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
CBSE
Subjects taught
Science, Mathematics, Hindi, Social Science, English
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Computer Science, Science, Physics, Hindi, Physical Education, English, French, EVS, Chemistry, Mathematics, Social science, Computers, History, Geography
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
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Mathematics, Social science, History, Science, English, Physical Education, Physics, EVS, Hindi, Geography, Computer Science, Computers, French, Chemistry
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
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Computers, Chemistry, French, EVS, History, Social science, Computer Science, English, Mathematics, Physics, Hindi, Science, Geography, Physical Education
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
3
Board
CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE), ICSE
Subjects taught
French, Social Science, Hindi, English, Science, Computers, Mathematics, Social studies, EVS
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
3
Board
ISC/ICSE, CBSE, IGCSE, International Baccalaureate, State
Subjects taught
Mathematics, Chemistry
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
3
Board
CBSE
Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Answered on 28/10/2018 Learn CBSE - Class 11/Chemistry
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Post a Lesson
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO3 i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO3 = 100/73 × 0.6844 g = 0.938 g
Like 0 
Answers 1 
Comments Answered on 25/10/2018 Learn Hydrocarbons
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Relative amount of monochlorinated product = Number of hydrogen X relative reactivity For (1°) monochlorinated product = 9x1 = 9
(2°) monochlorinated product = 2x 3.8 = 7.6 (3°) monochlorinated
product = 1x3 = 5
Total amount of monochlorinated compounds = 9 + 7.6+ 5 = 21.6
% of 1° monochlorinated product = 9x100/21.6 =41.67
% of 2° monochlorinated product =7.6x100/21.6 =35.18
% of 3° monochlorinated product =5x100/21.6 =23.15
Like 0 Answers 1 Comments Ask a Question
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Social Science, Hindi, English, Science, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
CBSE
Subjects taught
Science, Mathematics, Hindi, Social Science, English
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Computer Science, Science, Physics, Hindi, Physical Education, English, French, EVS, Chemistry, Mathematics, Social science, Computers, History, Geography
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
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Mathematics, Social science, History, Science, English, Physical Education, Physics, EVS, Hindi, Geography, Computer Science, Computers, French, Chemistry
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
3
Board
ICSE, CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Computers, Chemistry, French, EVS, History, Social science, Computer Science, English, Mathematics, Physics, Hindi, Science, Geography, Physical Education
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
3
Board
CBSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE), ICSE
Subjects taught
French, Social Science, Hindi, English, Science, Computers, Mathematics, Social studies, EVS
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
3
Board
ISC/ICSE, CBSE, IGCSE, International Baccalaureate, State
Subjects taught
Mathematics, Chemistry
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
3
Board
CBSE
Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Answered on 28/10/2018 Learn CBSE - Class 11/Chemistry
Ask a Question
Post a Lesson
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO3 i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO3 = 100/73 × 0.6844 g = 0.938 g
Like 0 Answers 1 Comments Answered on 25/10/2018 Learn Hydrocarbons
Ask a Question
Post a Lesson
Relative amount of monochlorinated product = Number of hydrogen X relative reactivity For (1°) monochlorinated product = 9x1 = 9
(2°) monochlorinated product = 2x 3.8 = 7.6 (3°) monochlorinated
product = 1x3 = 5
Total amount of monochlorinated compounds = 9 + 7.6+ 5 = 21.6
% of 1° monochlorinated product = 9x100/21.6 =41.67
% of 2° monochlorinated product =7.6x100/21.6 =35.18
% of 3° monochlorinated product =5x100/21.6 =23.15
Like 0 Answers 1 Comments Ask a Question
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Hyderabad
