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2 years ago

9

Addition of 6 m hcl to which substance will not result in gas evolution?

1 answer:

saul85 [17]2 years ago

6 0

There is no gaseous product when NaNO3 reacts with 6M HCl.

We must note that gas evolution is one of the signs that a chemical reaction has taken place. Hence, we must consider the options and look out for an option in which reaction with HCl does not lead to a gaseous product. This is the option that does not lead to evolution of a gas.

Let us consider the reaction; NaNO3(s) + HCl(aq) ---> NaCl(s) + HNO3(aq), there is no gaseous product hence NaNO3 does not result in gas evolution upon addition of 6M HCl.

Missing parts;

Addition of 6 M HCl to which substance will NOT result in gas evolution?

(A) Al

(B) Zn

(C) K2CO3

(D) NaNO3

Learn more about evolution of gas: brainly.com/question/2186340

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Explain why the solutions of SbCl3(aq) in our lab also contain HCl. Why would adding a few drops of this solution to 300 mL of w

IRISSAK [1]

Answer:

Here's what I get

Explanation:

SbCl₃ reacts with water to form slightly soluble antimony oxychloride.

SbCl₃(aq) +H₂O(ℓ) ⇌ SbOCl(s) + 2HCl(aq)

Your observation is an example of Le Châtelier's Principle in action,

The SbCl₃(aq) in your lab has enough HCl added to push the position of equilibrium to the left and keep the SbOCl in solution.

If a few drops of the SbCl₃(aq) were added to 300 mL of water, the solution would turn cloudy. The HCl would be so dilute that the position of equilibrium would lie to the right, and a cloudy precipitate of antimony oxychloride would form.

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3 years ago

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eduard

Answer:

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2 years ago

A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines

kodGreya [7K]

Answer : The rate law for the overall reaction is, $Rate=k[A][B]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $A+B\rightarrow AB$ (slow)

Step 2 : $A+AB\rightarrow A_2B$ (fast)

Overall reaction : $2A+B\rightarrow A_2B$

The rate law expression for overall reaction should be in terms of A and B.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

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The expression of rate law for this reaction will be,

$Rate=k[A][B]$

Hence, the rate law for the overall reaction is $Rate=k[A][B]$

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Empirical formula for compound of 2.17 mol N and 4.35 mol O

balu736 [363]

Answer:

Explanation:

ratio of moles of N and O in molecule =

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2 years ago

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