What is the systematic name for the following structure, specifying cis-/trans-?

Sucrose and naphthalene are both organic molecules. Did you find significant differences in their solubilities

pochemuha

Answer:

Naphtalene is very unlikely to get dissolved in water as it is nonpolar.

Sucrose is very likely to get dissolved in water as it is polar at the oxygen-to-hydrogen bonds.

Explanation:

Hello,

In this case, naphtalene (C₁₀H₈) is constituted by both carbon-to-carbon and carbon-to-hydrogen bonds only as it is an aromatic hydrocarbon which is highly nonpolar due to the low-polarity of the C-H bond, therefore, it is very unlikely to get dissolved in water. On the other hand, sucrose (C₁₂H₂₂O₁₁) has oxygen-to-hydrogen bonds which are highly polar, for that reason, it is able to get dissolved in water via hydrogen bonds.

Best regards!

4 0

3 years ago

Un método de laboratorio para preparar O 2 (g) consiste en la descomposición de KCIO 3 (s): KCIO 3 (s)  KCI(s) + O 2 (g) a) ¿Cu

FromTheMoon [43]

Answer:

2a. 0.401 moles de O2

b. 127.7g KClO3

c. 43.95g KCl

3. 9600g Ag2CO3 debieron haberse descompuesto

Explanation:

Basados en la reaccion:

2KCIO 3 (s) → 2KCI(s) + 3O 2

2 moles de KClO3 producen 3 moles de oxígeno:

a. 32.8g de KClO3 son:

Moles KClO3 -Masa molar: 122.55g/mol-

32.8g * (1mol / 122.55g) = 0.2676 moles KClO3

Moles O2:

0.2676 moles KClO3 * (3moles O2 / 2mol KClO3) =

0.401 moles de O2

b. 50g O2 son:

Moles O2 -Masa molar: 32g/mol-

50g O2 * (1mol/32g) = 1.5625 moles O2

Moles KClO3:

1.5625 moles O2 * (2mol KClO3 / 3mol O2) = 1.042 moles KClO3

Masa KClO3:

1.042 moles KClO3 * (122.55g/mol) =

127.7g KClO3

c. Las moles de O2 son:

28.3g O2 * (1mol / 32g) = 0.88 moles O2

Moles KCl:

0.88 moles O2 * (2mol KCl / 3mol O2) = 0.5896 moles KCl

Masa KCl -Masa molar: 74.55g/mol-

0.5896 moles KCl * (74,55g/mol) = 43.95g KCl

3. Basados en la reacción:

2Ag 2 CO 3 → 4Ag(s) + 2CO 2 (g) + O 2 (g)?

2 moles de Ag2CO3 reaccionan produciendo 4 moles de Ag. Las moles de Ag son:

Moles Ag -Masa molar: 107.8682g/mol-

7511g Ag * (1mol/107.8682g) = 69.63 moles Ag

Moles Ag2CO3:

69.63 moles Ag * (2moles Ag2CO3/4molesAg) = 34.82 moles Ag2CO3

Masa Ag2CO3-Masa molar: 275.7453g/mol-

34.82 moles Ag2CO3 * (275.7453g/mol) =

9600g Ag2CO3 debieron haberse descompuesto

4 0

3 years ago

How many molecules of carbon dioxide, CO2, are present in 388.1 grams?

swat32

Answer:

53.11× 10²³ molecules

Explanation:

Given data:

Number of molecules of CO₂ = ?

Mass of CO₂ = 388.1 g

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of CO₂ = 12× 1 + 16×2

Molar mass of CO₂ = 44 g/mol

Now we will put the values in formula.

Number of moles = 388.1 g/ 44 g/mol

Number of moles = 8.82 moles

Now we will calculate the number of molecules by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

8.82 mol × 6.022 × 10²³ molecules / 1 mol

53.11× 10²³ molecules

6 0

3 years ago

How many MOLES of boron tribromide are present in 3.20 grams of this compound ?

Dmitriy789 [7]

Answer:

1) There are 0.0128 moles of BBr3 present in 3.20 grams of this compound

2) There are 909.35 grams of BBr3 present in 3.63 moles of this compound

Explanation:

Step 1: Given data

Boron tribromide = BBr3

Molar mass of Boron = 10.81 g/mole

Molar mass of Bromide = 79.9 g/mole

Molar mass of Boron tribromide = 10.81 + 3*79.9 = 250.51 g/mole

Step 2: Calculating number of moles

Number of moles = mass / molar mass

Number of moles of BBr3 = 3.20 grams / 250.51 g/mole

Number of moles of BBr3 = 0.0128 moles

Step 3: Calculating mass

If we have 3.63 moles of boron tribromide (= BBr3)

Mass of BBr3 = number of moles of BBr3 * Molar mass of BBr3

Mass of BBr3 = 3.63 moles * 250.51 g/mole

Mass of BBr3 = 909.35 grams

7 0

3 years ago

How old is a fossil that has radioactively decayed through 4 half-lives of carbon-14?

gavmur [86]

Answer:

Carbon-14 has a half life of 5730 years. This means it takes 5730 years for half of the carbon-14 atoms to decay to carbon-12 in a given sample (eg: 100 carbon-14, one half life = 50 carbon-14, 50 carbon-12).

Explanation:

8 0

3 years ago

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