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Ulleksa [173]
3 years ago
5

What is the definition of lead in chemistry

Chemistry
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

Lead is a chemical element with the symbol Pb (from the Latin plumbum) and atomic number 82. It is a heavy metal that is denser than most common materials. Lead is soft and malleable, and also has a relatively low melting point.

Explanation:

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the answer should be six months

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Henok is observing the reaction rate of a chemical reaction. He then adds a catalyst to the reaction.
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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.

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4 0
1 year ago
8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0
3 years ago
Solve the problem. Express your answer to the correct number of significant figures
Shalnov [3]

Answer:

Explanation:

30.53131

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