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balu736 [363]
3 years ago
15

According to the law of conservation of mass, what should the product side of this single replacement reaction look like? A. H2

ZnCl2 B. H2 + Zn + Cl2 C. H2 + ZnCl2 D. ZnCl2
Chemistry
1 answer:
Katyanochek1 [597]3 years ago
8 0

<u>Answer:</u> The product side must be ZnCl_2+H_2

<u>Explanation:</u>

Single displacement reaction is defined as the reaction in which more reactive metal displaces a less reactive metal from its chemical reaction.

AB+C\rightarrow CB+A

Metal C is more reactive than metal A.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.

When zinc metal reacts with hydrochloric acid, it leads to the production of zinc chloride and hydrogen gas. The chemical reaction follows:

Zn+2HCl\rightarrow ZnCl_2+H_2

<u>On reactant side:</u>

Number of zinc atoms = 1

Number of hydrogen atoms = 2

Number of chlorine atoms = 2

<u>On product side:</u>

Number of zinc atoms = 1

Number of hydrogen atoms = 2

Number of chlorine atoms = 2

Hence, the product side must be ZnCl_2+H_2

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The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo
notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0
3 years ago
In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio
Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.  

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.  

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)  


Substituting these values we get                                                            

The total amount of caffeine obtained during each extraction                

 = (4.0×C )+ (2.0×4.6 × C)                                                                              

= 13.2 C


The amount of caffeine remaining in the aqueous solution is calculated as  

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)


Substituting these values we get                                                            

The amount of caffeine remaining in the aqueous solution = 4 × C                                                                                            

The fraction of caffeine remaining in aqueous solution is calculated as  

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)                    

=4.0 C/13.2 C                                                                                                

= 1/3.3.  

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3  =0.028

Therefore, the total amount of caffeine extracted                            

=0.070 × (1-(1/3.3)^3)                                                                                      

= 0.068 g


5 0
3 years ago
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Of the metals pb hg na and mg which will not spontaneously donate electrons to copper in solution
marta [7]

Answer : Hg (Mercury)


Explanation : In the given series of elements Na falls after the Cu in the reactivity series, for Pb it also falls after Cu, and for Mg it is the same.


Only Hg which is mercury can spontaneously donate its electrons to copper in the solution because it falls before Cu in the reactivity/activity series.

3 0
3 years ago
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What is the pH of pure water
Vlad1618 [11]

Answer: The answer is 7

Explanation:

6 0
3 years ago
PLSS HELP ASAP it's due
BartSMP [9]
<h3>Answer:</h3>

87.02%

<h3>Explanation:</h3>

Percent yield of a product in a chemical equation is the ratio of actual or experimental yield to theoretical yield expressed as percentage.

In this case we are given;

Mass of Zinc as 7.23 g

Actual volume of Hydrogen gas produced as 2.16 L

We are required to calculate the percentage yield of Hydrogen gas;

<h3>Step 1: Write a balanced equation for the reaction</h3>

The balanced equation for the reaction between Zinc metal and Hydrochloric acid is given by;

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

<h3 /><h3>Step 2: Moles of Zinc metal that reacted </h3>

Moles are given by dividing mass with molar mass

Moles = Mass ÷ Molar mass

Molar mass of Zinc = 65.38 g/mol

Therefore;

Number of moles = 7.23 g ÷ 65.38 g/mol

                             = 0.1106 moles

<h3>Step 3: Calculate the number of moles of Hydrogen gas produced</h3>

From the equation 1 mole of Zinc results in the formation of 1 mole of Hydrogen gas.

Therefore, Moles of hydrogen gas = Moles of Zinc × 1

                                                         = 0.1106 moles × 1

                                                         = 0.1106 moles

<h3>Step 4: Calculate the theoretical volume of Hydrogen gas produced.</h3>

At STP, 1 mole of a gas occupies 22.4 Liters

Therefore;

Volume of Hydrogen = Number of moles × 22.4 L

                                   = 0.1106 mole × 22.4 L

                                  = 2.477 L

<h3>Step 5: Calculate the percent yield </h3>

Percent yield = (Actual yield ÷ Theoretical Yield) × 100%

                      = (2.16 L ÷ 2.477 L)× 100%

                     = 87.20%

Thus, the percent yield of hydrogen gas produced is 87.02%

8 0
3 years ago
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