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Kipish [7]
2 years ago
11

(1) (a) In an equilibrium reaction between gases Q2 and R2 to form QR, the energy content of the reactants is 100KJ and that of

the product is 54KJ. The energy content of the activated complex is 210KJ. (i) Draw an energy profile diagram for the reaction. (ii) Determine the: I. activation energy of the reaction. II. enthalpy change A H of the reaction. (iii) Write a balanced equation for the reaction. (iv) State whether the reaction is exothermic or endothermic. (v) Give a reason for the answer given in 1(a)(iv). (b) (i) If sodium reacts with cold water, state ONE observation that would be made as the reaction Proceeds. (ii) Write a balanced chemical equation for the reaction in 1(a)(i) (iii) Calculate the volume of gas produced at s.t.p if 3.0g of sodium reacts with excess water. [Molar volume of gas at s.t.p = 22.4dm³; H = 1.0, O=16.0, Na = 23.0] (c) Name ONE allotrope of oxygen.​
Chemistry
1 answer:
andrezito [222]2 years ago
7 0

Answer:

a is my answer

Explanation:

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Calculate the concentrations of h2so3, hso−3, so2−3, h3o+ and oh− in 0.025 m h2so3.
sammy [17]
We will use this two reaction equation:

H2SO3 + H2O ↔ H3O+  +  HSO3-    Ka1 = 1.3 x 10^-2

HSO3-  + H2O ↔ H3O+   + SO3 2-    Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

              H2SO3 + H2O ↔ H3O+ +  HSO3- 

initial     0.025                        0            0

change   -X                             +X          +X

Equ       (0.025-X)                     X             X 

 
Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X
                   
 ∴[H3O+] = 0.0127 M


and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

                 = 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

            = 8.66 x 10^-13 M

- by using the ICE table for the second equation:

              HSO3- + H2O ↔ H3O+         + SO3 2-

initial    0.0127                      0.0127            0

change    -X                            +X                +X

Equ      (0.0127-X)                (0.0127+X)        X


when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X) 

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X 

∴[SO32-] = 6.3 x 10^-8
3 0
3 years ago
The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
3 years ago
In ironmaking, iron metal can be separated from iron ore (Fe2O3) by heating the ore in a blast furnace in the presence of coke,
mel-nik [20]

The limiting reactant is iron ore, the theoretical yield of iron metal is 701.344 kg, and the theoretical yield of carbon dioxide is 413.292 kg.

<h3>Stoichiometric problem</h3>

From the equation of the reaction:

2 Fe_2O_3(s) + 3 C(s) --- > 4 Fe(s) + 3 CO_2(g)

The mole ratio of iron ore to carbon is 2:3.

Mole of 1000 kg of iron ore = 1000000/159.69

                                          = 6,262 moles

Mole of 120 kg carbon = 120000/12

                                 = 10,000 moles

Thus, it appears that the carbon is in excess while the iron ore is limited in availability.

The mole ratio of the iron ore and the iron produced is 1:2. Thus, the equivalent number of moles of iron produced will be:

              6,262 x 2 = 12,524 moles

Mass of 12,524 moles of iron = 12,524 x 56

                                                = 701,344 g or 701.344 kg

Thus, the theoretical yield of iron is 701.344 kg.

The mole ratio of the iron ore and the carbon dioxide produced is 2:3. The equivalent mole of carbon dioxide produced will be:

         6,262 x 3/2 = 9,393 moles

Mass of 9,393 moles carbon dioxide = 9,393 x 44

                                                         = 413,292 or 413.292 kg

The theoretical yield of carbon dioxide is, therefore, 413.292 kg.

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

       

3 0
1 year ago
Every element has a unique amount of
tamaranim1 [39]
Protons is the answer
5 0
3 years ago
What is the SI unit for the mass of subatomic particles?
gogolik [260]

<em>hey, im jordan :)</em>

the SI unit for the mass of subatomic particles is <u>amu (atomic mass unit)</u>

<em>hope this helps!</em>

<em>have a great day :D</em>

4 0
3 years ago
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