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ivolga24 [154]
4 years ago
9

Would it be C..? More NO2 and SO2 would form..help..please and thank you

Chemistry
1 answer:
Scorpion4ik [409]4 years ago
7 0

Answer:

C. More NO2 and SO2 will form

Explanation:

Le Chatelier's Principle : It  predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc

It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.

In given Reaction SO3 is introduced(increased) .

So equilibrium will shift in the direction where SO3 should be consumed(decreased)

Hence the equilibrium will go in backward direction , i.e

NO + SO_{3} \rightarrow NO_{2} + SO_{2}

So more and more Of NO2 and SO2 will form

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This problem is providing us with a statement in which we need to figure out the word fitting in the blank. At the end, after analyzing the information, the word turns out to be colligative as show below:

<h3>Colligative properties.</h3>

In chemistry, colligative properties of solutions account for the behavior of a solution with respect to the pure solvent, to which a solute is added.

Among them, we have boiling point elevation, freezing point depression, vapor pressure lowering and osmotic pressure, which are all affected by the concentration of the solute but not by the identity of the solute.

In such a way, we conclude that the correct word that fits in the blank is colligative as shown below:

"Colligative properties depend on the concentration of a solute in a solution but not on the identity of the solute."

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Each species is a separate type of organism.

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3 years ago
How many molecules of iodine are produced when 9.3×1026 molecules of chlorine gas react with lithium iodide?
Bas_tet [7]

Answer:

9.3x10^{26}molec\ I_2

Explanation:

Hello!

In this case, considering the balanced chemical reaction:

Cl₂ + 2LiI ⇒ 2LiCl + I₂

We can see there is a 1:1 mole ratio between the produced iodine and the used chlorine, thus, we infer that the number of molecules of iodine given those of chlorine turn out:

9.3x10^{26}molec\ Cl_2*\frac{6.022x10^{23}molec\ I_2}{6.022x10^{23}molec\ Cl_2} =9.3x10^{26}molec\ I_2

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3 years ago
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