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snow_lady [41]
2 years ago
13

Based on the type or types of intermolecular forces, predict the substance in each pair that has the higher boiling point Match

the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer Reset Help stronger dispersion forcesG Given the molecules propane (C3Hs) and n-butane (C4H10), Г- has the higher boiling point mainly due to n-butane (C,H Given the molecules diethyl ether (CH3 CH2OCH2 CHs) and 1-butanol (CH3 CH2CH2 CH2OH) higher boiling point mainly due to 1-butanel CH,CH, CH,CH,OH) diethyl ether (CH CH,OCH CH, has the propane (C, H ts greater melar mass
Chemistry
1 answer:
rosijanka [135]2 years ago
5 0

Answer:

C4H10 has a higher boiling point due to stronger dispersion forces

1-butanol has the greater boiling point mainly due to hydrogen bonding influences

Explanation:

If we consider propane and n-butane, we discover that they are both alkanes. However, the magnitude of dispersion forces in alkanes depends on the length of the carbon chain. The greater the length of the carbon chain is greater in n-butane than in propane. The greater the chain length, the greater the magnitude of dispersion forces and the greater the boiling point. Hence, n-butane has a higher boiling point than than propane.

Also, if we compare the boiling points of 1-butanol to that of diethyl ether, we will discover that 1-butanol has a higher boiling point mainly due to intermolecular hydrogen bonding influences.

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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

3 0
3 years ago
500.0 mL of a gas is collected at 745.0 mmHg. What will the<br> volume be at 760 mmHg?
Natalija [7]

Answer:

490 ml

Explanation:

5 0
2 years ago
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
A booth renter is responsible for conducting all necessary business blank
tatyana61 [14]

Answer:

building and managing their own clients, purchasing supplies and keeping records

hope this helps you

5 0
3 years ago
The pH of orange juice is 5, what is the pOH?
Kay [80]
I think its 14 - 5 and you get 9
3 0
3 years ago
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