suggest a suitable electrolyte that could be used if the iron ring is to be plated with silver​

Answer:

The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:

CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)

Explanation:

Cobalt (ii) hydroxide is an inorganic compound consisting of cobalt (ii) ions, Co²+ and hydroxide ions, OH-. It is insoluble in water and the pure form known as the beta form is a pink-coloured solid. The impure form which incorporates other anions in its molecular structure is blue in colour and is ustable.

Cobalt (ii) hydroxide is formed as precipitate when an alkaline metallic hydroxide such as sodium hydroxide is mixed with an aqueous cobalt (ii) salt such as cobalt (ii) sulfate. The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:

CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)

Being a basic hydroxide, cobalt (ii) hydroxide neutralizes acids to form cobalt (ii) salts and water. For example: Co(OH)₂ (s) + H₂SO₄ (aq) ---> CoSO₄ (aq) + H₂O

Thus, cobalt (ii) hydroxide is soluble in acids.

Cobalt(II) hydroxide is used mostly as a drying agent for paints, varnishes, and inks. It is also useful in the preparation of other cobalt compounds.

Answer:

1. Multimeter

2. Cathode

3. Capacitor

4.  Selector switch

5. Short or Shorted

6. Resistors

7. Resistor color coding

8. Open Capacitor

9. Step-up transformer

10. Step-down transformer

Explanation:

The above-described elements are electronic components. Resistors for instance are designed to resist the flow of electric current. They are also standardized such that a deviation from the set resistance level will indicate a problem. The capacitor is another electrical component that stores energy as an electrical charge. Knowledge of these electrical components and the ways they are tested will make a person proficient in electrical electronics.

Answer:

he a real G

Explanation:

og tripple og

Answer:

i believe the answer is D: 18

Explanation:

Answer:

18

M shell can actually hold 18 electron as you move higher atomic number

Answer:

Wearing gloves

Explanation:

it helps on not touching chemicals

Answer:

Light can be treated like particles because it is made of chunks like things called protons.

Answer:

D

Explanation:

Answer: The mass of carbon dioxide required is 308 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of octane = 100.0 g

Molar mass of octane = 114.23 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of octane}=\frac{100.0g}{114.23g/mol}=0.875mol[/tex]

For the given chemical reaction:

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

By the stoichiometry of the reaction:

2 moles of octane produces 16 moles of carbon dioxide

So, 0.875 moles of octane will produce = [tex]\frac{16}{2}\times 0.875=7mol[/tex] of carbon dioxide

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

[tex]\text{Mass of carbon dioxide}=(7mol\times 44g/mol)=308g[/tex]

Hence, the mass of carbon dioxide required is 308 g

Answer:

Explanation:

From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]

This relates to the 1st order reaction rate, whereby:

The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]

here:

[A] = reactant concentration at time (t)

[A]_o = initial concentration for the reactant

k = rate constant

As such, the order of the reaction is the first order

Rate constant [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]

Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]

The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]

From the given table:

the initial concentration of [tex]H_2O_2[/tex] = 1.00 M

∴

We can determine the concentration of the reactant at 4000s by using the formula:

[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]

[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]

[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]

Finally, at 4000s: the average rate is:

[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

In the first reaction, a neutron is released, and in the second a helium atom is released. The given two equations represent nuclear fission and fusion.

A nuclear reaction is a reaction that involves the nuclei of the atom and the absorption and release of energy. In the first reaction, a big nucleus is split by the neutron bombardment into smaller nuclei.

In the second reaction the process of nuclear fusion, two nuclei combine into a single larger nucleus that is shown as:

₁¹H+ ²₁H → ³₂He

Therefore, nuclear fission and fusion are represented by each of these equations.

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Answer:

5–bromo–9–chlorodecane

Explanation:

To name the compound given above, the following must be obtained:

1. The longest continuous carbon chain. This gives the parent name of the compound.

2. The substituent group attached to the compound.

3. Position of the substituent group.

4. Combine the above to obtain the name.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.

2. The substituent groups attached to the compound are:

I. Bromine (Br) => Bromo

II. Chlorine (Cl) => Chloro

3. The position of the substituent groups are:

I. Br => carbon 5

II. Cl => carbon 9

NOTE: numbering is done alphabetically.

4. Therefore, the name of the compound is:

5–bromo–9–chlorodecane

Answer:

A.

Explanation:

I chose this answer and it was correct ‍♀️

Answer:

The solute is salt and the solvent is water

Explain:

Because salt is a component the dissolves in the solvent

Answer:

[tex]molar \: mass \: of \: copper(ii)nitrate = 64 + (14 + 48) \times 3 \\ = 250 \: g \\ 64 \: g \: of \: copper \: produces \: 250 \: g \: of \: copper \: nitrate \\ 10.36 \: g \: of \: copper \: will \: produce \: ( \frac{10.36 \times 250}{64} ) \: g \\ = 40.7 \: g[/tex]

Answer:

□In field investigations, you are usually working with much larger animals than in the lab. In field investigations, it is much more difficult to separate your control and experimental groups.

There are many ways in which a field  experiment can differ form of a classroom or a laboratory experiment,  one of which can be the environmental condition.

 A field experiment Is an experiment  which is performed in practically Outside the classroom or a laboratory in live situtations .

Environmental factors may deter the results  of a field experiment or  may alter the physical conditions of the object, For example  an anhydrous substance may become  hygroscopic, Which may  Alter the result of the field experiment.

  Hence,  field experiment can differ from  a classroom or a laboratory experiment  due to environmental conditions .

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Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Answer:

I don't no answer sorry

Explanation:

you follow me

The question is incomplete, the complete question is;

1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?

A. CL2

B. SF6

C. Kr

D. UF6

E. Xe

Answer:

SF6

Explanation:

From Graham's law;

Let the rate of diffusion of oxygen be R1

Let the rate of diffusion of unknown base be R2

Let the molar mass of oxygen by M1

Let the molar mass of unknown gas be M2

Hence;

R1/R2 = √M2/M1

So;

2.14/1 = √M2/32

(2.14/1)^2 = M/32

M= (2.14/1)^2 × 32

M= 146.6

This is the molar mass of SF6 hence the answer above.

Answer:

4

Explanation:

because it has 3 carbons  and 6 hydrogen

hope this helps :)

Answer:

Chromosomes and I think its too many

Explanation:

Answer:

Br

Explanation:

because bromine is more reactive as reactivity increases on moving from left to right in p-block. hope this make sense :)

Answer:

A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)

2. Strong acid (sa); Hydrochloric acid: HCl (aq)

3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)

B. H+ (aq) + OH- (aq) ----> H₂O (l)

Explanation:

Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.

The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:

1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)

Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions

2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)

Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions

Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:

3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)

Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.

Part B:

The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.

The net ionic equation of the neutralization reaction is given below :

H+ (aq) + OH- (aq) ----> H₂O (l)

Answer:

[tex]C_3H_7O_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:

[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]

Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:

[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]

Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:

[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]

To get:

[tex]CH_{2.34}O[/tex]

Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:

[tex]C_3H_7O_3[/tex]

Regards!

Explanation:

All atoms of one type were identical in mass and properties.

Answer:

Explanation:

25.71 × 10²³ molecules are there in 4.27 mol of tungsten (VI) oxide, [tex]WO_3[/tex].

To find the Number of molecules of substance multiply the number of moles of given substance by Avogadro number

Number of molecules = Number of moles × Avogadro's Number

Avogadro's number [tex](N_{A})[/tex] is the number of particles in one mole of substance. 6.023 × 10²³ is known as Avogadro's constant / Avogadro's number.

Avogadro's number = 6.023 × 10²³ [tex]mol^{-}[/tex]

Now,

Number of molecules in 4.27 moles of [tex]WO_{3}[/tex]

= Number of moles of [tex]WO_{3}[/tex] × Avogadro's Number

= 4.27 × 6.023 × 10²³

= 25.71 × 10²³ molecules

Thus, we can say that 25.71 × 10²³ molecules are there in 4.27 moles of tungsten (VI) oxide, [tex]WO_3[/tex].

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Answer:

P4 + 4NaOH + 2H2O → 2PH3 + 2Na2HPO3

Explanation:

A chemical equation is said to be balanced if the quantity of each type of atom in the reaction is the same on both the reactant and product sides. In a balanced chemical equation, the mass and the charge are both equal.

A chemical equation must balance according to the rule of conservation of mass. According to the rule, mass cannot be generated or removed during a chemical process.

Chemical equations must be balanced, which means that the atom types and numbers on both sides of the reaction arrow must match. Coefficients are the values added in front of formulas to balance equations; they multiply each atom in a formula.

Here the given equation is balanced as:

P₄ + 4NaOH + 2H₂O——> 2PH₃ + 2Na₂HPO₃

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Answer:

2

Explanation:

You write your answer with the same number of significant figures as the number with the smallest amount of figures, in this case, that number is 2.1 so your answer should be written with 2 significant figures.

Equilibrium shifts to the right.

OPTION A

Answer:

A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.

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