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dem82 [27]
2 years ago
7

An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

Chemistry
1 answer:
uysha [10]2 years ago
8 0

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

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6 0
3 years ago
The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3
bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻     pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq)   pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64

6 0
3 years ago
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3 years ago
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Answer:

The answer is: 0,13 moles of CO2

Explanation:

We use the formula PV=nRT. The conditions STP are 1 atm of pressure and 273K of temperature:

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3 years ago
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