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3 years ago

6. 100 ml of gaseous hydrocarbon consumes 300

Chemistry

1 answer:

mario62 [17]3 years ago

7 0

Answer:

a. C₂H₄

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the complete combustion reaction that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

a. C₂H₄:

C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

b. C₂H₂:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

с. С₃Н₈

C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

d. C₂H₆

2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

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E ground pushes you forward. But that interaction is friction. Reduce friction and it doesn't matter how strong your legs are, the surface is incapable of pushing you accordingly. The coefficient of static friction is very low so it is easy to slide your foot rather than push.

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2 years ago

Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2

NemiM [27]

Answer:

CuSO4

Explanation:

Na2S + CuSO4 → Na2SO4 + CuS

The reaction is balanced (same number of elements in each side)

To determine limiting reagent you need to know the moles you have of each.

Molar mass Na2S = 23 * 2 + 32 = 78

Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5

Na2S mole = 15.5 / 78 = 0.2

CuSO4 mole = 12.1/159.5 = 0.076

*Remember mole = mass / MM

With that information now you have to divide each moles by its respective stoichiometric coefficient

Na2S stoichiometric coefficient : 1

Na2S : 0.2 / 1 = 0.2

CuSO4 stoichiometric coefficient: 1

CuSO4: 0.076 / 1 = 0.076

The smaller number between them its the limiting reagent, CuSO4

8 0

3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$