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Alik [6]
3 years ago
13

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

(l) the value of δs° for this reaction is ________ j/k⋅mol.
Chemistry
1 answer:
meriva3 years ago
3 0

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}}

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Answer:

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Explanation:

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3 years ago
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The toxicant absorbtion can be reduced after exposure to the skin the surrounding clothing shoes or gloves should be removed or torn off than the part of the body which was exposured to the toxicant should be immediatly washed using clean running water for a while, with cold shower being the most recommended splashing method. In case of absorption for orally consumed chemicals should check on any remaining toxicant residue and be removed from the mouth. Vomiting should be induced to patients that are still conscious by providing them with liquids that can provoke vomiting. This will help in removing the toxicant in the intestinal and reduce their effect. Gastric lavage should then be done to induce diarrhea.  

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4 0
3 years ago
In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.
Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Li

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Given: For N_2

Given mass = 2.50 g

Molar mass of N_2 = 28.02 g/mol

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According to the given reaction:

6Li+N_2\rightarrow 2Li_3N

6 moles of Li react with 1 mole of N_2

1 mole of Li react with 1/6 mole of N_2

0.3602 mole of Li react with \frac {1}{6}\times 0.3602 mole of N_2

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Available moles of N_2 = 0.08924 moles

N_2 is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of Li_3N

1 mole of Li gives 2/6 mole of Li_3N

0.3602 mole of Li react with \frac {2}{6}\times 0.3602 mole of Li_3N

Moles of Li_3N = 0.12

Molar mass of Li_3N = 34.83 g/mol

Mass of Li_3N = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

<u>Theoretical yield = 4.18 g</u>

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3 years ago
What is the freezing point (in °C) of a 0.743 m
LUCKY_DIMON [66]

Answer:

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