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3 years ago

How would adding the catalyst nitrogen monoxide (NO) affect this reaction?

Chemistry

1 answer:

RSB [31]3 years ago

8 0

Answer: A) NO increases the rate at which $SO_3$ molecules are formed.

Explanation:

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more reactant molecules can cross the energy barrier by undergoing collisions and convert to products.

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

Thus NO will increase the rate of reaction by lowering the activation energy and thus the colllisions among $SO_2$ and $O_2$ molecules will incraese which in turn will lead to formatioon of more $SO_3$ molecules.

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Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

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Which of the following is an anion?Al3+ Mg2+ O2– H Save

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3 years ago

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Answer:

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3 years ago

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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

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4 years ago

A3. Answer each of the following: A student dissolved 1.3g of Ba(OH)2 (molar mass = 171.34 g/mol) in 250 mL of water and reacted

Sunny_sXe [5.5K]

Answer:

(i) Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O

(ii) 121.392 mL of HNO3 0.125M are required to react completely with the Ba(OH)2 solution.

(iii) The molarity of the Ba(OH)2 solution is 0.0303 M

(iv) Bromothymol Blue (pH range 6.0 - 7.6)

(v) pH of the soultion would be 2.446

Explanation:

(i) First of all, to solve this problem we should write the balanced chemical equation to know the stoichiometry of the reaction:

Ba(OH)2 + HNO3 → Ba(NO3)2 + H2O

The previous reaction simply describes the reactants and products involved in the chemical process. As you can see, the mass balance is not balanced because the quantity of atoms in the reactants side of the equation is not equal to the ones in the products side. So we try to add coefficients to the reaction in order to balance the amount of atoms on both sides of the reaction. To to this, we take a look at the reaction: We see that the main product formed Ba(NO3)2 has 2 atoms of N, so we add a number 2 besides the HNO3 to equal the quantity of Nitrogen atoms:

Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + H2O

Now, we can see that from the reactants side of the equation there are 8 atoms of Oxygen and in the products side we only have 7. Hence, we add the number 2 besides the molecule of water:

Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O

If we check the situation now, we can observe that all the atoms are balanced on both sides of the reaction, so We did it!

(ii) From the balanced equation we now know that 1 mole of Ba(OH)2 reacts with 2 moles of HNO3 to form the stated products. Let's see, therefore, how many moles of Ba(OH)2 are in solution:

According to the molar mass of Ba(OH)2: 1 mole = 171.34 g

So, the student add 1.3 g of the compound to water. This means that he added 7.587x10-3 moles of Ba(OH)2. This amount of Ba(OH)2 will react with 0.01517 moles of HNO3 taking into account the stoichiometry of the balanced equation described above (1 mol of Ba(OH)2 reacts with 2 moles of HNO3).

Now that we know the amount of moles of acid required to react with the hydroxide, we need to translate this moles into volume of acid solution:

We have a 0.125 M HNO3 solution. This means that there are 0.125 moles of HNO3 in 1000 ml of solution.

0.125 moles HNO3 ------ 1000 ml Solution

0.01517 moles --------- x = 121.392 ml HNO3 Solution

This means that we need 121.392 ml of a 0.125 M HNO3 solution to react completely with the Ba(OH)2 added by the student.

(iii) Now we are asked to calculate the molarity of the Ba(OH)2 solution. From the calculations performed before in point (ii) we know that the hydroxide solution consisted of 7.587x10-3 moles of Ba(OH)2 and that this quantity of moles were in 250 mL of water. So:

250 ml Solution ----- 7.587 x10-3 moles Ba(OH)2

1000 ml Solution ----- x = 0.0303 M

(iv) Since Ba(OH)2 and HNO3 are both strong base and acid respectively, they react with each other completely to form the salt Ba(NO3)2 and water. Therefore, the pH of the solution when the reactions ends will be neutral or nearly neutral (pH = 7) and because of this we need an indicator that would change its color around this pH to be able to visualize the end point of the titration. The Bromothymol blue serves this perfectly since its change in color ranges between pH 6.0 and 7.6.

(v) If we now calculate how many moles of HNO3 are present in 150 mL of a 0.125 M solution we obtain:

1000 mL solution ---- 0.125 moles HNO3

150 mL solution ------ x = 0.01875 moles.

From this, we know that if we add 150 mL of the acid solution we would have 0.01875 moles of HNO3. However, from the previous points, we know that 0.01517 moles of the compound will be consumed by the reaction with Ba(OH)2 leaving in solution only 3.58 x10-3 moles of HNO3 (0.01875 moles - 0.01517 moles).

This amount of HNO3 will dissociate according to the following equation:

HNO3 → H+ + NO3-

The amount of protons present in solution will determine the pH. Because, as we said before, Nitric acid is a strong acid, it will dissociate completely intro protons and nitrate. As a result of this, we would have 3.58 x10-3 moles of H+ in the solution (1 mole of HNO3 produces 1 mole of H+) and considering the contribution of protons in the solution given by the dissociation of the water negligible, then:

pH = - log [H+]

pH = - log [3.58 x10-3] = 2.446

3 0

3 years ago