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Sonja [21]
3 years ago
10

When a magnesium wire is dipped into a solution of lead (II) nitrate, a black deposit forms on the wire. Which of the following

can be concluded from this observation?
(A) The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
(B) Mg(s) is less easily oxidized than Pb(s).
(C) An external source of potential must have been supplied.
(D) The magnesium wire will be the cathode of a Mg/Pb cell.
(E) Pb(s) can spontaneously displace Mg2+(aq) from solution.
...

Chemistry
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).

Explanation:

Metals are usually arranged in an order of reactivity called activity series. Metals that are high up in the series are good reducing agents with very low (very negative) reduction potentials. Metals with greater (less negative) reduction potentials are found lower in the series. In the image attached, elements were arranged according to their reducing ability. Magnesium is very electro positive hence it is a better reducing agent with a lesser standard reduction potential than lead(refer to the image for numerical values of standard reduction potentials). Hence it displaces lead from solution and the elemental lead deposits on the wire.

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Salsk061 [2.6K]

Answer:

Explanation:

1. It is not Blanced. on the right the is 1S and 6O and the left there is 1S and 3O

2.If the energy level of the reactants is higher than the energy level of the products the reaction is exothermic (energy has been released during the reaction). If the energy level of the products is higher than the energy level of the reactants it is an endothermic reaction.

 

4 0
3 years ago
Consider the reaction: P4 + 6Cl2 = 4PCl3.
likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

                                P4      +      6Cl2      =      4PCl3

                          4(31) ---------- 12(35.5)

                         20     ----------    x

                    x = 20(12x35.5) / 4(31)

                   x = 8520 / 124

                   x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

                            P4      +      6Cl2      =      4PCl3

                       124g             426 g               4(31 + 3(35.5)) = 550g

                        15g               22g

I will use P4 to find the limiting reactant

                 

                     x = (15 x 426) / 124 = 51.5   The limiting reactant is Chlorine

                                                                  because we need 51.5 g and we only have 22g

Excess reactant

                 x = (22 x 124) / 426 = 6.4 g of P4

           Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

                              426 g of Cl2 ----------------  550 g of PCl3

                                 22g of Cl2 ------------- -     x

            x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

   % yield = (16.25  - 28.4) / 28.4 x 100

  % yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

                                   124g of P4  -------------      426 g  6Cl2

                                     28g           ---------------     x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3  - 96.2 = 10.1 g of Cl2 in excess

                   

                 

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Which conversion of energy always occurs in a voltaic cell?
kvasek [131]
(4) chemical energy to electrical energy is the correct answer.
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draw the organic product formed when 1−hexyne is treated with h2o, h2so4, and hgso4. click the draw structure button to launch t
V125BC [204]

The organic product formed when 1−hexyne is treated with H₂O, H₂SO₄, and HgSO₄ will be 2-hexanone (structure attached).

This reaction is an example of an oxymercuration reaction of the organic product 1−hexyne.

Oxymercuration is shown in three steps to the right. The nucleophilic double bond attacks the mercury ion, releasing an acetoxy group. The mercury ion's electron pair attacks carbon on the double bond, generating a positive-charged mercuronium ion. Mercury's dxz and 6s orbitals give electrons to the double bond's lowest unoccupied molecular orbitals.

In the second stage, the nucleophilic H₂O attacks the highly modified carbon, freeing its mercury-bonding electrons. Electrons neutralize mercury ions by collapsing. Water molecules have positive-charged oxygen.

In the third stage, the negatively charged acetoxy ion released in the first step attacks the hydrogen of the water group, generating the waste product HOAc. The two electrons in the oxygen-hydrogen link collapse into oxygen, neutralizing its charge and forming alcohol.

You can also learn about organic products from the following question:

brainly.com/question/13513481

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4 0
1 year ago
A decay series starts with the synthetic isotope ²³⁹₉₂U. The first four steps are emissions of a β⁻ particle, another β⁻, an a p
VladimirAG [237]

<u>Thorium series</u> could start by this sequence.

<h3>Brief explanation</h3>

To write balanced equations for nuclear decay processes. It's important to remember that the mass number and the atomic numbers must be balanced. And so what that means is that if we look at an elements nuclear symbol, the atomic number is the bottom number and the top number, the superscript, is the mass number, and so when we add them up on both sides, they have to be equal. There are two different ways in which decay can occur.

In this, series one is through beta decay, which means that the following particle is produced. The other is Alpha Decay, which produces this particle. Both are products. So if we start off with uranium to 39 you read it in nuclear notation, which means we have to find the atomic number just 92 and it undergoes beta decay.

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Except for that it's 93 equals minus one plus x, where X is 94 which is P u plutonium, and the mass number is zero plus X equals to 39 or to 39. The next decay starts with the isotope that we just form to 39 p. U. This time it's an Alpha decay. So we produce this particle to find the unknown. Element 94 equals two plus x, where X equals 92 which takes us back to uranium.

Find the mass number of this isotope 2 39 equals four plus X, where X equals to 35. Finally, for the last decay, you have another Alpha decay starting with uranium to 35 making an alpha particle. The atomic number will be 90 which is T H and the top is 2 31 For the mass number. This begins the natural decay, series of thorium .

Learn more about chemical decay

brainly.com/question/1898040

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7 0
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