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marta [7]
3 years ago
5

Manufacturing traditional clayware ceramics typically involves driving off the water of hydration in the clay minerals. The rate

constant for the dehydration of kaolinite, a common clay mineral, is 1×10−4s−1 at 485 ∘c and 1×10−3s−1 at 525 ∘c. Calculate the activation energy for the dehydration of kaolinite.
Chemistry
1 answer:
zimovet [89]3 years ago
3 0

The relation between activation energy, rate constant and temperature is given by Arrhenius equation

Arrhenius equation is

ln (K2/ K1) = Ea / R (1/T1- 1/T2)

K2 =  1×10−3s−1  T2 =  525 ∘C = 798 K

K1 = 1×10−4s−1  T1 =  485 ∘c = 758 K

Ea = ?

R = gas constant = 8.314 J / mol K

ln (K2/ K1) = ln (10^-3 / 10^-4) = 2.303 = Ea /8 .314 (1/758 - 1 / 798)

2.303 = Ea / 8.314 (0.00132 - 0.00125)

Ea = 2.303 X 8.314 / (0.00132 - 0.00125)  = 273530.6 J / mole

Ea = activation energy = 273.531 kJ / mole

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Answer:

The famous oil drop experiment exploits that fact that an oil drop in an electronic field will get negative charge accumulation which can be balanced and observed in order to determine the charge of an electron.
6 0
3 years ago
Read 2 more answers
What is the pOH of a 0.025 M [H+] solution? A.0.94 B.1.60 C.12.40 D.13.06
Greeley [361]

Answer:

option C= 12.40

Explanation:

Formula:

pH + pOH = 14

First of all we will calculate the pH.

pH = - log [H⁺]

pH = - log [0.025]

pH = - (-1.6)

pH = 1.6

Now we will put the values in formula,

pH + pOH = 14

pOH = 14-pH

pOH = 14 -1.6

pOH = 12.4

The pOH of solution is 12.4.

6 0
3 years ago
4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

To calculate the percentage yield of 3-hexanol, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0
3 years ago
Which of the following is an example of a homogeneous mixture?
elena-s [515]

A. Apple juice is an example of a homogeneous mixture.

7 0
3 years ago
Choose all true statements about the water molecule: Choose all true statements about the water molecule: Bonding between the ox
amm1812

Answer: Water is a permanent electric dipole, having permanent charge separation.

Explanation:

Hydrogen bonding is an intermolecular force having partial ionic-covalent character.

In H_2O, O is a highly electronegative atom attached to a H atom through a covalent bond. The oxygen atoms being more electronegative gets partial negative charge and H atom gets partial positive charge. Thus water is permanent electric dipole.

Hydrogen bonding takes place between a hydrogen atom (attached with an electronegative atom O) and an electronegative atom (O).

4 0
3 years ago
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