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3 years ago

2. What happens to the reaction rate if a catalyst is added?

Chemistry

2 answers:

Ghella [55]3 years ago

6 0

Answer:

A catalyst speeds up a chemical reaction

Explanation:

It increases the reaction rate by lowering the activation energy for a reaction.

mrs_skeptik [129]3 years ago

6 0

It will speed up the chemical reaction.

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You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25

ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of AgCN = n = $\frac{125 g}{134 g/mol}=0.9328 mol$

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

$V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL$

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

6 0

3 years ago

If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

igomit [66]

Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$

Best regards.

5 0

2 years ago

Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3

Margarita [4]

Answer: The rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

$2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O$

The intermediate reaction of the mechanism follows:

Step 1: $H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}$

Step 2: $H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}$

Step 3: $HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[H_3O_2^+][Br^-]$ ......(1)

As, $[H_3O_2^+]$ is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for $[H_3O_2^+]$ from step 1, we get:

$K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}$

$[H_3O_2^+]=K[H^+][H_2O_2]$

Putting the value of $[H_3O_2^+]$ in equation 1, we get:

$\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]$

Hence, the rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

4 0

3 years ago

What 3 gases does the flame test check for and what should you expect to see/hear?

Kazeer [188]

What of the gasses is N and the flame changes it's color from Orange to blue

7 0

3 years ago

Which of the following acts as a catalyst in catalytic converters?

Serhud [2]

Answer:

B. Metal

Explanation:

The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium. Platinum is used as a reduction catalyst and as an oxidation catalyst. Although platinum is a very active catalyst and widely used, it is very expensive and not suitable for all applications.

Hope it helps plz mark brainlist :)

5 0

3 years ago

Read 2 more answers