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2 years ago

What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

2 answers:

7 0

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

ivann1987 [24]2 years ago

7 0

Answer:

-30.02 ºC

Explanation:

Assuming the antifreeze to be ethylene glycol (C₂H₆O₂) which is popular antifreeze.

Molar mass of ethylene glycol (C₂H₆O₂) = 62g/mol

Step 1: calculate the freezing point depression of the solution

ΔT = -Kf*M

where,

ΔT= depression in the freezing point.

M = the molarity of the solution (mol solute / Kg solvent)

Kf = molar freezing point constant of water = 1.86°C/m

To determine depression in the freezing point (ΔT), first we need to calculate;

molarity of solute (ethylene glycol) in mol
mass of solvent (water) in kg
molarity of the solution (water +ethylene glycol)

Step 2: calculate the molarity of the solute (ethylene glycol)

Molar mass ethylene glycol = 62 g/mol

molarity of ethylene glycol in mol = 50 g / 62g/mol = 0.807 mol

Step 3: calculate mass of solvent in kg

There is 1kg of ethylene glycol which is present in 1kg of water

mass of solvent (water) in kg= 50 g/ 1000 g/ Kg = 0.050 Kg

Step 4: calculate the molarity of the solution (M)

M = 0.807 mol / 0.050 Kg = 16.14 m

Step 5: calculate the freezing point depression of the solution (ΔT)

ΔT = - Kf*M = -1.86 ºC/m x 16.14 m

= -30.02 ºC

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A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

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$0=120m*a-180m*a+36s^{2}*a^{2}$

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$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

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If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

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Since both forces are the same,

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The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

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