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nata0808 [166]
3 years ago
6

PLEASE HELP...with single replacement reaction w/ activity series

Chemistry
1 answer:
Blizzard [7]3 years ago
8 0

Answer : The balanced chemical reaction will be:

AuI_4+2Br2\rightarrow AuBr_4+2I_2

Explanation :

Single replacement reaction : A chemical reaction in which the more reactive element replace the less reactive element.

It is represented as,

A+BC\rightarrow AC+B

In this reaction, A is more reactive element and B is less reactive element.

As per question, when gold (IV) iodide react with bromine to give gold (IV) bromide and iodine.

The balanced chemical reaction will be:

AuI_4+2Br2\rightarrow AuBr_4+2I_2

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Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s
Brrunno [24]

She would observe a yellowish  solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.

This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing  particles of iodide. Upon mixing,the  lead particles  from the Lead nitrate solution combines with the  iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid  precipitate called Potassium nitrate.

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2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)

See similar answer here  :https://brainly.in/question/46262462

4 0
2 years ago
Read 2 more answers
If a frog initially contained 2 grams of carbon-14 and the half-life of carbon-14 is 5,730 years, how much carbon-14 remains in
andreyandreev [35.5K]
Half life is the time taken for a radioactive isotope to decay by half its original mass. In this case the half life of carbon-14 is 5.730 years. 
Using the formula;
New mass = original mass × (1/2)^n; where n is the number of half lives (in this case n=1 ) 
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5 0
3 years ago
Calculate the standard free-energy change at 25 ∘C for the following reaction:
lianna [129]

Answer:

Standard free-energy change at 25^{0}\textrm{C} is -3.80\times 10^{2}kJ/mol

Explanation:

Oxidation: Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)

Reduction: Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)

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

Overall: Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)

Standard cell potential, E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}

So, E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V

We know, standard free energy change at 25^{0}\textrm{C}(\Delta G^{0}): \Delta G^{0}=-nFE_{cell}^{0}

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, \Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol

8 0
3 years ago
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