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
Answer:
0.09 x10^-10m
Explanation:
Using wavelength=( 12.27 A)/√V
= 12.27 x 10^-10/ √1.6x10^2
= 0.09x10^-10m
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
Answer:
106.03 meters
Explanation:
The height is given by the formula for motion under the influence of gravity.
h = -4.9t^2 +162.7
Height is 0 when ...
0 = -4.9t^2 +162.7
4.9t^2 = 162.7
t^2 = 162.7/4.9
t = √(162.7/4.9)
The horizontal distance traveled in that time is ...
(18.4 m/s)√(162.7/4.9) s ≈ 106.03 m
The object will strike the ground about 106.03 meters from the base of the cliff.
Answer:
V initial = 29.4 m.s²
Explanation:
( Using the laws of motion)
V final = V initial + Acceleration × time
0 = V initial + ( -9.8)(3)
29.4 = V initial
* I took upward as positive that's why I substituted -9.8 *
* for V final we know that at maximum height the ball is not moving thats why is = 0 *
