What is the molarity of 20.2 g of potassium nitrate, KNO3, in enough water to make 250.0 mL of solution

Consider the following mechanism.

solong [7]

Answer :

(a) The overall equation is:

$ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)$

(b) The intermediates are :

$OH^-(aq),HClO(aq)\text{ and }HIO(aq)$

Explanation :

(1) $ClO^-(aq)+H_2O(l)\rightarrow HClO(aq)+OH^-(aq)$ (fast)

(2) $I^-(aq)+HClO(aq)\rightarrow HIO(aq)+Cl^-(aq)$ (slow)

(3) $OH^-(aq)+HIO(aq)\rightarrow H_2O(l)+IO^-(aq)$ (fast)

By adding the three equations and cancelling the common terms on both side, we will get the overall equation.

$ClO^-(aq)+I^-(aq)\rightarrow Cl^-(aq)+IO^-(aq)$

Intermediates are generated and consumed in the mechanism and do not include in the overall equation.

Since, intermediates will not include in the overall mechanism.

The intermediates are :

$OH^-(aq),HClO(aq)\text{ and }HIO(aq)$

7 0

3 years ago

Calculate the molarity of a solution made by adding 45.4 g of nano3 to a flask and dissolving it with water to create a total vo

aleksandr82 [10.1K]

Solutions are made up of two non reacting species called solute and solvent. The amount of solute in solvent is known as concentration of that solute. Concentration is often measured in Molarity. Molarity is the amount of solute dissolved in 1 dm3 of solution. Answer to your question is as follow;

3 0

3 years ago

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State one use for argon.

Alex17521 [72]

Argon is perticularly important for the metal industry, being used as an inert gas shield in arc

8 0

3 years ago

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Find the volume of a box with a length of 5 cm, a width pf 5cm, and a height of 10cm.

Ray Of Light [21]

You so you do 5x5x10 and that equals 100. Hope this helped(:

6 0

3 years ago

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25 points , hi! please look at the attachment for the question, I'm having a hard time because all I have to do is solve through

MAXImum [283]

Answer:

To answer the question, we correctly fill the attached screenshot as follows;

3H₂ + N₂ → 2NH₃
The molar mass of H₂ = 2 g/mol

$100.0 g \ H_2 \times \dfrac{1 \ mol \ H_2}{2 \ g \ H_2} \times \dfrac{2 \ mol \ NH_3}{3 \ mol \ H_2} = 33.\bar 3 \ mol \ NH_3$

The molar mass of N₂ = 28 g/mol

$100.0 g \ N_2 \times \dfrac{1 \ mol \ N_2}{28 \ g \ N_2} \times \dfrac{2 \ mol \ NH_3}{1 \ mol \ N_2} \approx 7.143 \ mol \ NH_3$

A. Therefore, the excess reactant is hydrogen gas H₂ because it makes the most amount of ammonia, NH₃ (33. $\bar 3$ moles of NH₃)

B. The limiting reactant in nitrogen, N₂, because it is the reactant that makes the least amount of the ammonia, NH₃ (approximately 7.143 mol NH₃)

C. The theoretical yield of ammonia, is the maximum amount of ammonium that can be produced from the reaction between the 100 g of hydrogen gas, H₂, and 100 g of nitrogen gas, N₂ which is given by the amount of ammonia produced by the limiting reactant which is approximately 7.143 mole of NH₃

Explanation:

7 0

3 years ago

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