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maks197457 [2]
3 years ago
15

1) Which of the following best describes the following compound: Cu3N2

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

Answer:

A is the answer

Explanation:

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 This is more to do with Bioaccumulation of mercury, where the mercury is absorbed into the issue of animals, and so animals higher in the food chain consume a lot of tissue matter hence increasing mercury content in their system.
7 0
3 years ago
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.
Alex

Answer:

Part A

Ag+ is the Lewis acid and NH3 is the Lewis base.

Part B

AlBr3 is the Lewis acid and NH3 is the Lewis base.

Part C

AlCl3 is the Lewis acid and Cl− is the Lewis base.

Explanation:

A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.

A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .

3 0
3 years ago
Which of the following is an example of an observation?
Bess [88]
THE ANSWER TO THE QUESTION IS A
8 0
3 years ago
Proportional means that
FromTheMoon [43]
Corresponding in size or amount to something else.
5 0
3 years ago
How many molecules are there in 560 grams of CoCl2?​
Marrrta [24]

Answer:

25.89  × 10²³ molecules

Explanation:

Given data:

Mass of CoCl₂ = 560 g

Number of molecules present = ?

Solution:

Number of moles of CoCl₂:

Number of moles = mass/molar mass

Number of moles = 560 g/ 129.84 g/mol

Number of moles = 4.3 mol

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

4.3 mol ×  6.022 × 10²³ molecules /1 mol

25.89  × 10²³ molecules

5 0
3 years ago
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