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Mnenie [13.5K]
3 years ago
14

Carbon diselenide (CSe2) is a liquid at room temperature. The normal boiling point is 125°C, and the melting point is –45.5°C. C

arbon disulfide (CS2) is also a liquid at room temperature with normal boiling and melting points of 46.5°C and –111.6°C, respectively. How do the strengths of the intermolecular forces may vary from CO2 to CS2 to CSe2?
Chemistry
1 answer:
Tema [17]3 years ago
5 0

Answer:

Explanation:  The strengths of the inter molecular forces varies as follows -

        CO_{2}< CS_{2} < CSe_{2}

The normal boiling point of CSe2 is 125°C and that of CS2 is 116°C, which explains the trend that as we move down the group, the boiling point of e compound increases as the size increases.

This usually happens because larger and heavier atoms have a tendency to exhibit greater inter molecular strengths due to the increase in size . As the size increases, the valence shell electrons move far away from the nucleus, thus has a greater tendency to attract the temporary dipoles.

And larger the inter molecular forces, more tightly the electrons will be held to each other and thus more thermal energy would be required to break the bonds between them.

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1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an
salantis [7]

Answer:

1) empirical formula = CH2O ; the molecular formula = C2H402

2) empirical formule = C3H404 = Molecular formula

Explanation:

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H

⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H =  5.3045 g of O

2) Calculate number of moles

C: 3.9736 g / 12.0 g/mol = 0.3311 mol

H: 0.6629 g / 1.0 g/mol = 0.6629 mol

O: 5.3045g / 16.0 g/mol =  0.3315 mol

3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements

C: 0.3311/ 0.3311= 1

H: 0.6629 / 0.3311 = 2

O: 0.3315/ 0.3311  = 1

This gives us the empirical formula of CH2O

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 12 + 2*1 + 16 = 30g

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

60 / 30 = 2 = n

If we multiply CH2O by 2 we'll get: C2H4O2

If we control this by calculating the molar mass:

2*12 + 4*1.01 + 2*16 = 60.04 g/mole

Then, the molecular formula is C2H4O2

CxHyOz  → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom  This is 2g H/ 16g O

In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H

⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H =  8.715g of O

2) Calculate number of moles

C: 4.90 g / 12.0 g/mol = 0.4083 mol

H: 0.545 g / 1.0 g/mol = 0.545 mol

O: 8.715g / 16.0 g/mol =  0.5447 mol

3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements

C: 0.4083/ 0.4083= 1

H: 0.545 / 0.4083 =1.33

O: 0.5447/ 0.4083  = 1.33

We should multiply everything by 3

This gives us the empirical formula of C3H4O4

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 3*12 + 4*1 + 4*16 = 104

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

104 / 104 = 1 = n

This means the empirical formula = molecular formula = C3H4O4

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3 years ago
If one serving of a packaged frozen dinner contains 5.2 grams of saturated fat and the maximum recommended daily intake of satur
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Answer:

The percentage of the daily value for saturated fat provided by one package of this food is 26.0%

Explanation:

It has been given from the preceding statement that one package of the dinner provides 5.2g out of the 20 grams recommended for daily need. This is 5.2/20 x 100 given 26.0% of the daily neded value will be provided by a plate or package of the dinner.

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Що таке кислоти? pliz<br>​
Shtirlitz [24]

Answer:

The last word means acid.

Explanation:

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What is the hybridization of bromine in BrO2-?
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<span>The hybridization of bromine must be sp^3.</span>
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What does CoH12O6 tell you about the glucose molecule?
Leokris [45]

Answer:

Glucose = C6H12O6

molecular mass = 6(12) + 12(1) + 6(16)

= 72 + 12 + 96

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Explanation:

Glucose has a chemical formula of: C6H12O6 That means glucose is made of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms. ... Glucose is produced during photosynthesis and acts as the fuel for many organisms.

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