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ElenaW [278]
2 years ago
7

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

ng system holds 4.90 gal, what is the boiling point of the solution?
Chemistry
1 answer:
iVinArrow [24]2 years ago
6 0

Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

           i is the van´t Hoff factor

           Kb the boiling constant which for water is 0.512 ºC/molal

           m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb =  Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

                                 = 9273.25 g / 62.07 g/mol =149.4mol

molality solution =  149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb =  Kbm = 0.512  ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

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1.4 mols

4th answer

Explanation:

22. 5 g of O2 in moles = (22.5/32) mols = 0.703 mol

The stoichiometry between O2 and H2O =1: 2

Therefore H2O produced = 2 * 0.703 mols=1.406 mols

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C. Kidneys filter wastes from the bloodstream and produce urine

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Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.
padilas [110]

The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

brainly.com/question/13996238

#SPJ4

5 0
1 year ago
3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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3 years ago
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