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LenKa [72]
3 years ago
6

HELP.....From the information presented in the activity series of metals below (4 metals included), answer the following questio

n.
In a single replacement reaction, would aluminum replace magnesium to form a new compound or would there be no reaction?

------------------------------------

magnesium (most reactive)

aluminum

zinc

iron (least)

a)there would be no reaction

b) aluminum would replace magnesium to form a new compound
Chemistry
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

a)there would be no reaction

Explanation:

The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.

In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.

 

Would aluminum replace magnesium to form a new compound or would there be no reaction?

Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.

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NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3?
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<h2>NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3? </h2><h2> </h2><h2>A. It has a very low pH. </h2><h2>B. It's dissociation is a reversible reaction. </h2><h2>C. It has a high H+ concentration. </h2><h2>D. It will release all of its OH- ions.</h2>

Explanation:

<h3>NH3 is a weak alkali that does not dissociate fully into its solution: It's dissociation is a reversible reaction. </h3><h3></h3>

Reactions are also :

  • Reversible
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Reversible reaction

A reaction in which products can combine back to give reactants under same given condition .

Example : N₂+H₂-------NH₃

Irreversible reaction

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Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
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Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

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