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rjkz [21]
2 years ago
4

A common antifreeze for car radiators is ethylene glycol, CH2(OH )CH2(OH ). How many millilite~s of this substance would you add

to 6.5 L of water 1n the radiator if the coldest day in winter is - 20°C? Would you keep this substance in the radiator in the summer to prevent the water from boiling? (The density and boiling point of ethylene glycol are 1.11 g cm^-3 and 470 K repsectively.)
Chemistry
1 answer:
scZoUnD [109]2 years ago
4 0

Explanation:

It is given that volume of water is 6.5 L. Therefore, mass of water present in 1 kg/L will be calculated as follows.

                  6.5 L \times 1 kg/L = 6.5 kg

As 1000 grams are present in 1 kg. So, 6.5 kg equals 6500 grams.

K_{f} of water is 1.86 K kg/mol. And, \Delta T = -20^{0}C

Let the weight of glycol is x.

It is known that \Delta T_{f} = K_{f} \times molality

Molar mass of ethylene glycol is 62 g/mol.

Hence, putting values in the above formula as follows.

                \Delta T_{f} = K_{f} \times molality

                  20 = 1.86 K kg/mol \times \frac{x \times 1000}{62 \times 6500}

                    x = 4333 g

As density equals mass present in per unit volume. Therefre, calculate the density as follows.

                 Density = \frac{mass}{volume}

                 volume = \frac{mass}{density}

                               = \frac{4333 g}{1.11 g cm^{-3}}

                               = 3903.60 cm^{-3}

                               = 3903 mL

The presence of solute will help in elevating the boiling point of solvent. Hence, theoretically it is possible to keep this substance in car radiator also in summer.

Whereas it is not necessary to add this glycol to water in summers as temperature does not reach the boiling point of water which is 373 K.

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icang [17]

Answer : 3.2 X 10^{15} g/cm^{3}

Explanation :  To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = 1.66054 X 10^{-24} g

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X  1.66054 X 10^{-24} g/amu = 1.673 X 10^{-24} g

Now, Volume = 1/6πd³ as diameter is given as 1.0 X 10^{-15} m converting it to cm will require to multiply with 100

∴ Volume  = 1/6π (1.0 X 10^{-15}mX 100 cm / 1 m)^{3}

Hence, volume =  5.236 X 10^{-40} cm^{3}

Therefore, Density = mass / volume

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6 0
2 years ago
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A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =
IgorC [24]
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
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by substitution with the new value of T(75+273) = 348K & A to get the new K
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2 years ago
When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro
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Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

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Hello there!

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