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
2 months agoContributor-Level 10
Molecular formula from calculation comes to be C3H6O i.e., it stands for the following compounds.

No addition reaction with Br2

VD * 2 = MM = 29 * 2 = 58
2.9 g of ether will combines with
Y = 9 and X = 8
Y - X = 9 - 8 = 1
New answer posted
2 months agoContributor-Level 10
From ΔG° = ΔH° – TΔS°
⇒ – 2.303RTlog Kc = ΔH° – TΔS° = 77.2 * 10³ – 400 * 122 = 28400 J
So log K? = (28400 / (2.303*8.314*400) ∴ [Kc = 1.958 * 10? ]
New answer posted
2 months agoContributor-Level 10
At Boyle's Temperature ; gas behaves ideally for a range of pressure.
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
- 65k Colleges
- 1.2k Exams
- 678k Reviews
- 1800k Answers




