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2 years ago

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Hexane is a common laboratory solvent and a component of petrol. Its combustion reaction is as follows: 2 C6H14 (9) + 19 02(9) +

12 CO2(g) + 14 H20(1) ArH=-8326 kJ mol-1 Use the reaction enthalpy value to calculate the heat in (kJ) released by combustion of 0.121 mol of hexane. (Give your answer as a positive number to at least THREE significant figures and do not include the units in the answer box)

1 answer:

Flauer [41]2 years ago

4 0

Answer : The heat released by the combustion of 1.121 mol of hexane (kJ) is, $1.01\times 10^3$

Explanation :

The given balanced reaction is,

$2C_6H_{14}(g)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(l)$

Given:

Moles of hexane = 0.121 mol

$\Delta H$ = -8326 kJ/mol

The negative sign indicate that the reaction exothermic or amount of heat released.

As, 1 mole of hexane released heat = 8326 kJ

So, 0.121 mole of hexane released heat = 8326 × 0.121 = 1007.446 ≈ $1.01\times 10^3kJ$

Therefore, the heat released by the combustion of 1.121 mol of hexane is $1.01\times 10^3kJ$ .

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Jim collects a sample of beach sand during a class trip. After a close inspection of the sample he classifies it as a ____________ mixture. A) heterogeneous

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draw the lewis structure for CO2, H2CO3, HCO3-, and CO3 2-.Rank these in order of increasing attraction to water molecules. Expl

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Answer:

The structures are attached in file.

Hydrogen bonding and intermolecular forces is the reason for ranks allotted.

Explanation:

In determining Lewis structure, we calculate the overall number of valence electrons available for bonding. Making carbon (the least electronegative atom) the central atom in the structure, we allocate valence electrons until each atom has achieved stability.

In order of decreasing affinity to water molecules:

$CO_{3}^{2-} > HCO_{3} ^{2-} > H_{2} CO_{3}$

This is due to the fact that the $CO_{3}^{2-}$ will accept protons more readily than the bicarbonate ion, $HCO_{3} ^{2-}$ . Carbonic acid, $H_{2} CO_{3}$ will not accept any more protons, hence it is the least attractive to water molecule, even though soluble.

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2 years ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

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2 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

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Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

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Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>( $K_{a}$ )<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

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<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

<u>The acid dissociation constant: </u> $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

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$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

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<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

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2 years ago

The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at

antoniya [11.8K]

Answer:

- 7.48

Explanation:

Given:

Concentration of the sugar solution, C = 0.3 M

Temperature, T = 27° C = 273 + 27 = 300 K

Now,

The solute potential is given as:

solute potential = - iCRT

where,

i is the number of particles the particular molecule will make in water

i = 1 for sugar

R is the universal gas constant = 0.0831 liter bar/mole-K

on substituting the respective values, we get

solute potential = - 1 × 0.3 × 0.0831 × 300

or

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2 years ago