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____ [38]
3 years ago
11

The reaction shown below has a positive enthalpy change and a negative entropy change. 2C (s) + 2H2 (g) yields C2H4 (g) Which of

the following best describes this reaction?
The reaction will be spontaneous at any temperature.
The reaction will not be spontaneous at any temperature.
The reaction will only be spontaneous at low temperatures.
The reaction will only be spontaneous at high temperatures.
Chemistry
1 answer:
Drupady [299]3 years ago
5 0
Answer is: <span>The reaction will not be spontaneous at any temperature.

</span>

<span>Gibbs free energy (G) determines if reaction will proceed spontaneously.
ΔG = ΔH - T·ΔS.
ΔG - changes in Gibbs free energy.
ΔH - changes in enthalpy.
ΔS - changes in entropy.
T is temperature in Kelvins.
When ΔS < 0 (negative entropy change) and ΔH > 0 (endothermic reaction), the process is never spontaneous (ΔG> 0).</span>

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3 years ago
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
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
3. The scientific study of how organisms are classified is called _____.
Vilka [71]
Hey mate!

Taxonomy is the <span>study of classification of organisms. Therefore, your answer is A.

Hope this helps!</span>
3 0
3 years ago
Jason is preparing curry from an Indian cookbook his British brother-in-law gee him for his birthday. Unfortunately, the British
Marina CMI [18]

Answer:

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Explanation:

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1 g equals 0,0022 pounds approximately

Then we need to cross-multiply:

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We can do the same calculation for the other 2 ingredients

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Quick electron emissions are called
Montano1993 [528]
Defined as a phenomenon of liberation of electron from the surface that is stimulated by temperature elevation, radiation, or by strong electric field.
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