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vaieri [72.5K]
2 years ago
7

The proof for the power property of logarithms appears in the table with an expression missing. Step reason mc027-1. Jpg given m

c027-2. Jpg substitution properties of exponents mc027-3. Jpg logarithm property mc027-4. Jpg mc027-5. Jpg substitution mc027-6. Jpg commutative property of multiplication which expression is missing from the proof?.
SAT
1 answer:
jolli1 [7]2 years ago
5 0

Proofs are used to confirm or disprove mathematical statements

The missing expression in the proof is : \log_b((b^{xy})

<h3>How to complete the proof</h3>

From the question, the first two steps are:

\log_b(M^y)

\log_b((b^x)^y)

By the properties of exponents, we have:

\log_b((b^{xy})

The above expression represents the missing expression in the proof of the logarithmic expression

Hence, the missing expression is : \log_b((b^{xy})

Read more about mathematical proofs at:

brainly.com/question/14036678

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Answer:

ummm is there anything else and whyyyyy? is this a math problem????? Probably. I think that many apples would very negatively impact his health.

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Determine the empirical formula of a compound containing 47. 37 grams of carbon, 10. 59 grams of hydrogen, and 42. 04 grams of o
Effectus [21]

Considering the definition of empirical formula, the empirical formula is C₃H₈O₂ and the molecular formula is C₉H₂₄O₆.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, you have:

  • Carbon (C): 47.37 grams
  • Hydrogen (H): 10.59 grams
  • Oxygen (O):  42.04 grams

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • Carbon (C): \frac{47.37 grams}{12 \frac{g}{mole} }= 3.95 moles
  • Hydrogen (H): \frac{10.59 grams}{1 \frac{g}{mole} }= 10.59 moles
  • Oxygen (O):  \frac{42.04 grams}{16 \frac{g}{mole} }= 2.63 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • Carbon (C): \frac{3.95 moles}{2.63 moles }= 1.5
  • Hydrogen (H): \frac{10.59 moles}{2.63 moles }= 4
  • Oxygen (O):  \frac{2.63 moles}{2.63 moles }= 1

Finally, since in the empirical formula the numbers of moles must be expressed in whole numbers, the ratio of atoms is multiplied by some number with which everything is obtained in simple whole numbers (in this case by 2):

  • Carbon (C): 1.5× 2= 3
  • Hydrogen (H): 4× 2= 8
  • Oxygen (O):  1× 2= 2

Therefore the C: H: O mole ratio is 3: 8: 2

Finally, the empirical formula is C₃H₈O₂.

<h3>Molecular formula</h3>

To obtain the molecular formula you must relate its molecular weight (PMc) with the molecular weight of the empirical formula (PMfe).

PMfe= PM (C₃H₈O₂)= 76 g/mole

The molar mass of the compound (PMc) was determined to be 228.276 g/mol. Dividing this mass PMc by the molar mass of the empirical formula PMfe gives:

228.276 g/mol ÷ 76 g/mole= 3

This means that in the molecular formula there are 3 unit formulas (empirical formulas), so it is necessary to multiply the number of all atoms by 3.

Then, the molecular formula is C₉H₂₄O₆.

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Relationship break up because of a number of reasons name and explain two factors.
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1. misunderstanding.

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Answer:

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