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3 years ago

{Plz answer} Thank you!

Chemistry

2 answers:

Dafna11 [192]3 years ago

8 0

The Answer to your question would be A

Umnica [9.8K]3 years ago

7 0

I would say A. Hope this helps, and cya! :3

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Calculate the number of gold atoms in a sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and r

Fudgin [204]

Answer: N = 2.78 × 10^23 atoms

There are N = 2.78 × 10^23 atoms in 70g of Au2cl6

Completed Question:

Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits

Explanation:

Given:

Molar mass of Au2cl6 = 303.33g/mol

Mass of Au2cl6 = 70g

Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol

According to the chemical formula of Au2cl6,

1 mole of Au2cl6 contains 2 moles of Au

Number of moles of Au = 2 × 0.231mol = 0.462mole

There are 6.022 × 10^23 atoms in 1 mole of an element.

Number of Atom of gold in 0.462 mole of gold is:

N = 0.462 mol × 6.022 × 10^23 atoms/mol

N = 2.78 × 10^23 atoms

7 0

2 years ago

Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom

ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

Average atomic weight of Hafnium= sum of all the atomic weights of the isotopes/ Total occuring isotopes

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

Average atomic weight= 1063.661amu /6 = 177.2768333amu

= 177.277amu to 3 decimal places.

6 0

2 years ago

Please help me with this basic organic conversion. use the shortest way to do it

jonny [76]

Answer:

See explanation

Explanation:

This conversion must go through a sequence of steps as i have shown in the image attached to this answer.

The acetone is converted to propan-2-ol using LiAlH4, THF and acid. The propan-2-ol may be converted to propene by E2 elimination. Addition of HBr yields 2-bromo propane.

The Wurtz reaction converts 2-bromo propane to 2,3- dimethyl butane. This can be brominated in the presence of light to yield 3-bromo-2,3-dimethyl butane. Elimination of HBr using a base leads to the formation of the required product as shown.