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2 years ago

The equilibrium for the indicator methyl orange is HC14H14SO3(red) + H2O C14H14SO3−(yellow) + H3O+ The reaction is exothermic.

Chemistry

1 answer:

muminat2 years ago

4 0

Answer:

The solution will turn red.

Explanation:

HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺

(red) (yellow)

Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,

At about pH 3.4, the two the forms are present in equal amounts, and the indicator colour is orange.

If you add more acid, you are disturbing the equilibrium.

According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.

The system will try to get rid of the added acid, so the position of equilibrium will move to the left.

More of the unionized molecules will form, so the solution will turn red.

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nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

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gizmo_the_mogwai [7]

Explanation:

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rosijanka [135]

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Let x be the number of gold atoms. Plug in the values to a calculator.

x = $\frac{5.00g}{3.271 x 10-22g}$
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Answer:

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Explanation:

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