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1 year ago

How could one rapidly separate red #40 from zinc oxide? indicate every step.

Chemistry

2 answers:

Lorico [155]1 year ago

8 0

The mixture of Red # 40 and zinc oxide is a heterogeneous mixture which can be separated by filtration

<h3>Further explanation</h3>

The mixture is a combination of two or more substances whose constituent properties do not change. The properties of the mixture component are not lost / unchanged as in compounds.

The mixture can be divided into a homogeneous mixture if the composition/ratio of each substance in the mixture is the same and heterogeneous mixture if the ratio of the composition of the substances is not the same in each place.

Mixtures can also be divided into solutions, suspensions, and colloids based mainly on the size of the particles

There are various mixed separation techniques including filtration, distillation, evaporation, extraction or chromatography

Red # 40, is an artificial coloring obtained from mixing hydrocarbons and is used in dyes such as candy or drinks

Zinc oxide or ZnO is an inorganic powder form which is used as an additive in the manufacture of products such as ceramics or glass and is not soluble in water but can dissolve in acids or bases

A heterogeneous mixture can be separated through physical separation techniques

The mixture of Red # 40 and zinc oxide is a heterogeneous mixture which can be separated by filtration because

Red # 40 can dissolve in water while zinc oxide is insoluble

Zinc oxide and red # 40 is filtered from the mixture using a filter paper

The filtering results called filtrate will be zinc oxide and the residual filtering called residue is Red # 40

While the separation of red # 40 and zinc oxide from water using evaporation techniques to obtain the dry solids

<h3><em>Learn more </em></h3>

A heterogeneous mixture

brainly.com/question/5428813

brainly.com/question/1080253

the difference between a solution and a heterogeneous mixture

brainly.com/question/5177216

Keywords: A heterogeneous mixture, red # 40, ZnO, filtration

tensa zangetsu [6.8K]1 year ago

6 0

Red #40 is soluble in water while zinc oxide is not.
So the easiest way to separate them is as follows:
1- add water to the mixture until red #40 is dissolved in the water
2- filter to separate the zinc oxide
4- heat the solution of red #40 and water until water evaporates and red#40 remains.

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A hydrogen atom is removed from the first carbon atom of a butane molecule and is replaced by a hydroxyl group. draw the new mol

Leno4ka [110]

The new molecule that is formed is an alcohol and is known as butanol. It has a chemical formula CH3CH2CH2CH2OH or C4H9<span>OH</span>. A molecule that has a hydroxyl functional group is an alcohol. It has isomers namely tert-butanol, 2-butanol and isobutanol.

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1 year ago

) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.

lidiya [134]

Answer:

- The order of the reaction based on the graph provided is first order.

- 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

8 0

1 year ago

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

Yuki888 [10]

<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$