Dilsukhnagar, Hyderabad, India - 500059.
Details verified of Ashish Singh✕
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Hindi Mother Tongue (Native)
English Proficient
Pondicherry University Pursuing
Bachelor of Science (B.Sc.)
Dilsukhnagar, Hyderabad, India - 500059
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Mathematics, Hindi, Science, Social Science, English
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
CBSE
Subjects taught
Hindi, English, Social Science, Mathematics, Science
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
History, Social science, Science, Computers, French, Geography, Mathematics, Hindi, Computer Science, EVS, Chemistry, Physics, Physical Education, English
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 7 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
Physics, History, Computers, Social science, Physical Education, Science, English, Hindi, French, Chemistry, EVS, Geography, Computer Science, Mathematics
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
EVS, Computers, Geography, English, Physics, Hindi, Mathematics, Computer Science, History, French, Chemistry, Physical Education, Science, Social science
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
3
Board
CBSE, ICSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Social studies, EVS, Mathematics, Computers, Science, Social Science, Hindi, French, English
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
3
Board
International Baccalaureate, IGCSE, State, CBSE, ISC/ICSE
Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
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
1. Which school boards of Class 10 do you teach for?
CBSE
2. Do you have any prior teaching experience?
Yes
3. Which classes do you teach?
I teach 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 2 years.
Answered on 28/10/2018 Learn CBSE/Class 11/Science/Chemistry
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
Answered on 25/10/2018 Learn Hydrocarbons
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
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Mathematics, Hindi, Science, Social Science, English
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
2
Board
CBSE
Subjects taught
Hindi, English, Social Science, Mathematics, Science
Taught in School or College
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
History, Social science, Science, Computers, French, Geography, Mathematics, Hindi, Computer Science, EVS, Chemistry, Physics, Physical Education, English
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 7 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
Physics, History, Computers, Social science, Physical Education, Science, English, Hindi, French, Chemistry, EVS, Geography, Computer Science, Mathematics
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
3
Board
Cambridge Assessment International Education (CAIE), CBSE, International Baccalaureate, ICSE, State
Subjects taught
EVS, Computers, Geography, English, Physics, Hindi, Mathematics, Computer Science, History, French, Chemistry, Physical Education, Science, Social science
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
3
Board
CBSE, ICSE, State, International Baccalaureate, Cambridge Assessment International Education (CAIE)
Subjects taught
Social studies, EVS, Mathematics, Computers, Science, Social Science, Hindi, French, English
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
3
Board
International Baccalaureate, IGCSE, State, CBSE, ISC/ICSE
Subjects taught
Mathematics, Chemistry
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
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/Science/Chemistry
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
Answered on 25/10/2018 Learn Hydrocarbons
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
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Dilsukhnagar, Hyderabad 
