The safety equipment must ………………. Inside the lab at all time brainly.

A humpback whale is classified as a mammal while a basking shark is a fish.

kvasek [131]

The taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

<h3>What is Taxonomic Classification?</h3>

We want to find Which taxonomy classification are the same for a humpback whale and a basking shark.

A chordate in the animal kingdom is referred to as Kingdom Animalia of the phylum Chordata. This phylum is one that includes all the vertebrates and subphyla tunicates and cephalochordates.

Chordates have peculiar characteristics such as notochord, pharyngeal slits, nerve cord, and a tail.

Therefore, we can conclude that the taxonomic classification that is the same for a humpback whale and a basking shark is Phylum.

Read more about Taxonomic Classification at; brainly.com/question/21023329

8 0

2 years ago

If the price of oil increases by 50% but consumption decreases by 25%, then the demand for oil is elastic.

abruzzese [7]

False is the right answe choose it

7 0

2 years ago

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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3 0

2 years ago