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3 years ago

A dilation produces a smaller figure. Which is a possible scale factor? –2 one-third 1 4.

SAT

1 answer:

Sedbober [7]3 years ago

7 0

Answer:

we know that

if a dilation produces a smaller figure

then

0< scale factor <1

case A) -2------> the scale factor must be a positive number

case B) 1/3----> 0< 1/3 <1---> is a possible scale factor

case C) 1------> this scale factor does not not produce a smaller figure, it produces the same figure

case D) 4----> this scale factor produces a larger figure

the answer is

1/3

Explanation:

please mark branlyist if correct

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san4es73 [151]

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2 years ago

Watching television for a long time is not good for you...........

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Answer:

It can cause headaches and pains at a close distance. It depends how big the monitor / TV.

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3 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