Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
Answer:
C. Neutral
Explanation:
Ions will combine in a way that the overall ionic compound will always be neutral.
Answer:
71 rpm
Explanation:
Given that:
Angular momentum (L) = 0.26
Diameter = 25cm = 0.25 cm
Radius, r = (d/2) = 0.125m
Mass = 5.6 kg
Moment of inertia (I) = 2mr² / 5
I = (2 * 5.6 * 0.125^2) / 5
= 0.175
= 0.175 / 5
= 0.035 kgm²
Angular speed (w) ;
w = L / I
w = 0.26 / 0.035
= 7.4285714
= 7.429 rad/s
w = (7.429 * 60/2π)
w = 445.74 / 2π rpm
w = 70.941724
Angular speed = 70.94 rpm
= 71 rpm
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
