Class 11th
Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
3 months agoContributor-Level 10
for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y) ∈ R
x³ - 3x²y - xy² + 3y³ = 0
⇒ (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
⇒ (3x - y) (x² - y²) = 0
Not symmetric
New answer posted
3 months agoContributor-Level 10
MnO? + H? C? O?2H? O - (H? )-> Mn²? + CO?
nf = (5) nf = (2)
Milli equivalent of C? O? ²? = mili equivalent of MnO?
2 x M x 10 = 5 x 0.05 x 10
M = 0.125 M
M = Strength / M.M of H? C? O?2H? O
Strength = 0.125 x 126g/L
= 15.75g/L
New answer posted
3 months agoContributor-Level 10
H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)
New answer posted
3 months agoContributor-Level 10
E.C = σ1s² σ1s² σ2s² σ2s² σ2p? ² π2p? ² = π2p? ² π2p? ² = π2p? ²
Total electrons in BMO = 10
New answer posted
3 months agoContributor-Level 10
C (graphite) → C (diamond)
ΔH? = Δ? H° (graphite) - Δ? H° (Diamond) = -832.8 - [-834.8] = 2kJ/mole
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 681k Reviews
- 1800k Answers


