What is the Lewis structure for NbBr5

Chemistry

1 answer:

Artist 52 [7]3 years ago

3 0

Answer:

Explanation:

add lone pairs on the outer bromines.

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At what temperature <br> does water boil

Ipatiy [6.2K]

The temperature is a 100

5 0

3 years ago

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There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine

SSSSS [86.1K]

1) You need to use the atomic mass of copper.

You can find it in a periodic table. It is 63.546 amu.

2) The atomic mass is the weigthed mass of the different isotopes.

This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:

=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.

3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x

=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278

=> 63.546 = 62.9296x + 64.9278 - 64.9278x

=> 64.9278x - 62.9296 = 64.9278 - 63.546

=> 1.9982x = 1.3818

=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%

=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%

Answer:

Cu-63 69.15%;

Cu-65 : 30.85%

3 0

3 years ago

Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me

LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps