What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the density of ethanol is 0.71gcm3

Answers

Answer 1

Answer:

Explanation:

Expression for capillary rise is as follows

h = 2T / ρ g r where T is surface tension , ρ is density of liquid and r is radius of capillary tube.

T = .032 J m⁻²

ρ = .71 g / cm³

= 7100 kg / m³

r = .1 x 10⁻³ m = 10⁻⁴ m

h = 2 x .032 / (7100 x 9.8 x 10⁻⁴ )

h = .0092 m

= 9.2 mm .


Related Questions

Identify which trends in the diagrams below describe
atomic radius, ionization energy, electron affinity,
and electronegativity

Answers

Explanation:

the atmoc radius across the period

Atom radius as you move form left to right the atomic in the period table atomic radius decrease do to decrease the number of shell and it causes the Electron closed to nucleus. and also cause to decrease the inozation engey, electronegative andelectro affinity....... in the group the atomic radius increase as you move down the group do to increase number of shell because the Electron are far-away to the nuclear and it causes the small amount of energy to ionization so that the ionization engey , electronaffity and electronegative are increase

The half-life of the radioactive isotope polonium-218 is 3.05 minutes.

How long will it take for the activity of a sample of polonium-218 to decrease from 2.81E4 Ci to 3.52E3 Ci?

Answers

Answer: It will take 9.13 minutes for the sample.

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  

t = age of sample

a = initial amount of the reactant = [tex]2.81\times 10^4[/tex]

a - x = amount left after decay process = [tex]3.52\times 10^3[/tex]

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]k=\frac{0.693}{3.05min}=0.227min^{-1}[/tex]

b) for activity to decrease from 2.81E4 Ci to 3.52E3 Ci:

[tex]t=\frac{2.303}{0.227}\log\frac{2.81\times 10^4}{3.52\times 10^3}[/tex]

[tex]t=9.13min[/tex]

Thus it will take 9.13 minutes for the sample.

[tex]t_{99.9}=40min[/tex]

The time after which 99.9% reactions gets completed is 40 minutes

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