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taurus [48]
3 years ago
12

Calculate the kinetic energy of co2 at 254 k .

Chemistry
1 answer:
san4es73 [151]3 years ago
3 0
You need to take into account the equations for the average speed and kinetic energy of gases.

Kinetic energy, KE = [1/2] m * v^2

Average speed, v = √[3RT/m] => v^2 = 3RT / m

=> KE = [1/2]m * 3RT/m = [3/2] RT

R = 8,314 J/mol*K

T = 254 K

=> KE = [3/2]*(8,314J/mol*K)*(254K) = 3167.6 J/mol

Answer: 3167.6 J/mol
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A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c
gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat  capacity and the temperature change. Working with this equation, and assuming no heat is lost to  the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat.  </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write  the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

3 0
3 years ago
What can pressure be changed by
Mariana [72]

Pressure is affected by gravity,height, and density since pressure = density×height×gravity .

6 0
3 years ago
Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.
Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

<em>Where [] are concentrations in equilibrium</em>

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized  over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>
8 0
3 years ago
If you have a density of 3g/ml and volume of 10mL.what is the mass?
ddd [48]
Density = Mass/Volume => Mass = Density*Volume => Mass = 3g/ml*10ml =30g
4 0
3 years ago
Read 2 more answers
Hydrochloric acid, or HCl, reacts with solid NaOH
yuradex [85]

Answer:

NaCI and H20


To simplify it.

5 0
3 years ago
Read 2 more answers
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