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3 years ago

Which statement best describes what an ocean tide is?

Chemistry

2 answers:

elena55 [62]3 years ago

8 0

Answer: B

Explanation: high tides and low tides are usually caused by the moon. The moons gravitational pull generates something called the tidal force. I’m not gonna go deep into this but yeah I think the answer is B

anyanavicka [17]3 years ago

3 0

Answer:

Explanation:

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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

A lab technician mixes a 0.730 M solution of sodium bromide (NaBr) and water. The volume of the solution is 135 milliliters.

Ivenika [448]

Answer:

We need 10.14 grams of sodium bromide to make a 0.730 M solution

Explanation:

Step 1: Data given

Molarity of the sodium bromide (NaBr) = 0.730 M

Volume of the sodium bromide solution = 135 mL = 0.135 L

Molar mass sodium bromide (NaBr) = 102.89 g/mol

Step 2: Calculate moles NaBr

Moles NaBr = Molarity NaBr * volume NaBr

Moles NaBr = 0.730 M * 0.135 L

Moles NaBr = 0.09855 moles

Step 3: Calculate mass of NaBr

Mass NaBr = 0.09855 moles * 102.89 g/mol

Mass NaBr = 10.14 grams

We need 10.14 grams of sodium bromide to make a 0.730 M solution

3 0

3 years ago

Read 2 more answers

Examine the molecular structure of this lipid. What type of lipid is this?

artcher [175]

The correct answer is Phospholipid

3 0

3 years ago

4.Identify each substance in the following word equation as a reactant or a product

Anastaziya [24]

Explanation:

4. limestone heat lime + carbon dioxide

The reactants in this expression above is limestone

The products of the reaction is carbon dioxide and lime

Reactant is the species that gives the product and it is usually found on the left hand side of the expression.

The product is the substance on the right hand side of the expression that forms through the experiment.

Heat is used to facilitate the reaction.

5. An exothermic reaction is a reaction in which heat is given off.

An endothermic reaction is a reaction in which heat is absorbed in the process.

An exothermic reaction is always warmer after the reaction whereas an endothermic reaction is colder at the end of the reaction.

5 0

3 years ago

An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well Exponents in rate laws are

Karo-lina-s [1.5K]

Answer:

- False.

- True.

Explanation:

Hello, for each statement we state:

- An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well.

FALSE because considering a rate law like:

$-r=kC^2$

The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

$-r=kC$

The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

- Exponents in rate laws are based on the coefficients from the balanced equation.

FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

- The rate constant, k, takes into account the effect of activation energy and temperature on the reaction.

TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

$k=Aexp(-\frac{Ea}{RT})$

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

$\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...$

Best regards.

5 0

3 years ago