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n200080 [17]
3 years ago
12

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de

termine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. 390. k 412 k 368 k 466 k 255 k
Physics
1 answer:
atroni [7]3 years ago
5 0
Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in
kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}.

The velocity is the first time-derivative of <em>r(t).</em>

<em />v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}<em />

The acceleration is the first time-derivative of the velocity.

a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}

Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,

a = -16\hat{j}

Its magnitude is 16 m/s².

4 0
2 years ago
A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released
GREYUIT [131]

Answer:

6.75\mu C/m^2

Explanation:

We are given that

Diameter,d=1\mu m=1\time 10^{-6} m

1\mu m=10^{-6} m

Radius,r=\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m

Density,\rho=900kg/m^3

Total number of electrons,n=39

Charge on electron =1.6\times 10^{-19} C

Total charge=q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C

Distance,s=2mm=2\times 10^{-3} m

Mass =density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg

Initial velocity,u=0

Final speed,v=4.5 m/s

v^2-u^2=2as

(4.5)^2-0=2a(2\times 10^{-3})

20.25=4a\times 10^{-3}

a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2

Force,F=ma

qE=ma

q(\frac{\sigma}{2\epsilon_0})=ma

\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}

\epsilon_0=8.85\times 10^{-12}

\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2

6 0
3 years ago
Est ce qu’un cricket c’est un consommateur
Sholpan [36]

Answer:

Oui

Explanation:

7 0
3 years ago
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f =  μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0
3 years ago
Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength
9966 [12]
Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

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Rarefraction - spot on a compressional wave where the waves are farther apart.
3 0
3 years ago
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