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Lesechka [4]
3 years ago
14

Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures

. For example, both butane and isobutane have the same molecular formula: C4H10- Calculate the mole percent of isobutane in an equilibrium mixture at 25° C, given that the standard Gibbs energy of formation of butane is -15.9 kJ/mol and that of isobutane is -18.0 kJ/mol. Does your result support the notion that straight-chain hydrocarbons (that is, hydrocarbons in which the C atoms are joined in a line) are less stable than branched-chain hydrocarbons?
Chemistry
1 answer:
kolezko [41]3 years ago
4 0

Answer:

Explanation:

Butane ⇒ Isobutane

ΔG = RT ln K

ΔG = - 18 + 15.9 = - 2.1 kJ / mol

- 2.1 x 10³ = 8.3 x 298 x lnK

lnK = .85

K = 2.33

[ isobutane ] / [ butane ]  = 2.33

isobutane / total mixture = 2.33 / 3.33 = .7

% of isobutane = 70 % .

70% is isobutane and 30 %  butane . Since equilibrium is tilted towards isobutane , it is more stable . Butane is straight chain and isobutane is branched chain compound .

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Acid discharge can enter the groundwater system and render it unfit for use.

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3 years ago
during a neutralization reaction phosphoric acid reacts with potassium hydroxide to produce potassium phosphate and water. if th
Otrada [13]

During a neutralization reaction phosphoric acid reacts with potassium hydroxide to produce potassium phosphate and water, the reaction occurs with a 64.9% yield.

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Potassium phosphate and water are created when potassium hydroxide and phosphoric acid combine. The reaction's balanced equation is as follows: K3PO4(aq) + 3H2O = 3KOH(aq) + H3PO4(aq) (l) If six moles of potassium hydroxide react, phosphoric acid is used in the reaction.

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2 years ago
The density of dry air measured at 25 Celsius is 1.19 times 10 to the negative 3 what is the volume of 50.0g of air
Oksana_A [137]

Answer:

4.2 x 10⁴ mL

Explanation:

Data Given:

Density (d) of air =  1.19 x 10⁻³g/mL

Mass of the air (m) = 50 g

Volume of the air (V) = ?

Solution:

Formula will be used

                d = m/V

As we have to find volume so rearrange the above equation

               V = m/d . . . . . . . . . . . (1)

Put values in above equation 1

                V = 50 g / 1.19 x 10⁻³g/mL

                V = 4.2 x 10⁴ mL

So,

volume of dry air = 4.2 x 10⁴ mL

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