States of Matter

Get insights from 92 questions on States of Matter, answered by students, alumni, and experts. You may also ask and answer any question you like about States of Matter

Follow Ask Question
92

Questions

0

Discussions

4

Active Users

0

Followers

New answer posted

a month ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Z = P V m R T 0 . 5 = 2 4 * V m 2 4

V m = 0 . 5 l / m o l e

( P + a V m 2 ) ( V m b ) = R T

( 2 4 + a 0 . 5 2 ) ( 0 . 5 0 . 1 ) = 2 4

2 4 + a 0 . 5 2 = 6 0

a 0 . 5 2 = 3 6

a = 3 6 * 0 . 5 2 = 9 a t m l 2 m o l 2
 

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

1 st   H - atom

n = 4

n = 1

4 3,3 2,2 1

  4 1 or 4 3,3 1

or 4 2,2 1

Hence, number of different lines possible = 3 + 1 = 4

Minimally, both can have same transition i.e 4 1 etc.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

From ΔT? = K? * m, and ΔT? = 15°C
∴ m = ΔT? / K? = 15 / 1.86 = 8.06
So the amount of propyl alcohol to be added.
= m * molwt = 8.06 * 60 = 483.6 g

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

At Boyle's Temperature ; gas behaves ideally for a range of pressure.

New answer posted

a month ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

P? = 1atm
P? = 1atm * 40/100 = 0.40atm
V? = 100 cm³
V? =?
At constant temperature, P? V? = P? V?
So V? = (P? V? ) / P? = (1atm * 100 cm³) / 0.40 atm = 250cm³
Hence, the volume of bulb B = (250 –100) cm³ = 150 cm³

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

16 [Z] = 16 [3/8] = 6

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Ease of liquefaction depends on a/b value.

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

PV = nRT
P = nRT . (1/V)
Plot of P vs (1/V) would be straight line passing through origin having slope = nRT.
At high temperatures, P vs (1/V) would have greater slope.

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Covalent bonding is NOT an intermolecular force while rest all are considered as intermolecular forces.

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

n (B) = 3/2
n (B) = 3, n (O) = 2
n (Total) = 3 + 2 = 5
X (B) = n (B) / n (T) = 3/5
X (O) = n (O) / n (T) = 2/5
P (S) = P° (B)X (B) + P° (O)X (O)
P (S) = 280 * (3/5) + 420 * (2/5)
= 336 mm of Hg

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 682k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.