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3 years ago

What are the products of the complete combustion of a hydrocarbon in excess oxygen

1 answer:

4 0

The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.

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silver has an atomic mass of 107.868 amu. silver has two common isotopes. one of the isotopes has a mass of 106.906 amu and a re

Sunny_sXe [5.5K]

A. The abundance of the 2nd isotope is 48.119%

B. The mass of the 2nd isotope is 108.905 amu

Let the 1st isotope be A

Let the 2nd isotope be B

A. Determination of the abundance of the 2nd isotope

Abundance of isotope A = 51.881%.

<h3>Abundance of isotope B =? </h3>

Abundance of B = 100 – A

Abundance of B = 100 – 51.881

<h3>Abundance of B = 48.119%</h3>

B. Determination of the mass of the 2nd isotope

Atomic mass of silver = 107.868 amu.

Mass of 1st isotope (A) = 106.906 amu

Abundance of isotope A (A%) = 51.881%.

Abundance of isotope B (B%) = 48.119%

<h3>Mass of 2nd isotope (B) =? </h3>

$atomic \: mass = \frac{mass \: of \:A \times \:A\%}{100} + \frac{mass \: of \:B \times \:B\%}{100} \\ \\ 107.868 = \frac{106.906\times \ \: 51.881}{100} + \frac{mass \: of \:B \times \:48.119}{100} \\ \\ 107.868 = \: 55.464 + 0.48119 \times mass \: of \:B \\ \\ collect \: like \: terms \\ \\ 0.48119 \times mass \: of \:B = 107.868 - 55.464 \\ \\ divide \: both \: side \: by \: 0.48119 \\ \\ mass \: of \:B = \frac{107.868 - 55.464 }{0.48119} \\ \\ mass \: of \:B =108.905 \: amu \\ \\$

Therefore, the mass of the 2nd isotope is 108.905 amu

Learn more: brainly.com/question/7955048

6 0

2 years ago

Complete these structures by adding electron dots as needed.

LiRa [457]

This is hard to show but here is how you would determine these. NOTE each dot is an electron.
Question 1) 
F-H 
1) determine the valance electrons for each. F has 7 and H has 1 
2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F 
.. 
F-H 
¨ 
Question 2) 

2) H-O-H 
H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place 
H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place 
O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place 

H-O-H: place two dots above and below the oxygen 

Question 3) 
3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each 
O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place 
N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place 
H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place 
place the 2 dots on top and bottom of oxygen. 
place 2 above the N

7 0

3 years ago

What is the frequency of radiation.

Ivenika [448]

RADIATION. Radio waves, microwaves, IR, light, UV, x-rays, GAMMA waves etc are ALL Electromagnetic radiation. The difference between ALL the above is the frequency, I.E. The number of waves per second. The higher the frequency the more energy.

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3 years ago

Hi friends!!!!!!!!!!!

Anna11 [10]

Answer:

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2 years ago

Read 2 more answers

Please help me with number 4 and 5 on my study guide

Shtirlitz [24]

Answer:

help with problem 4

Explanation:

4. protons are equal to atomic numbers. neutrons are atomic mass minus atomic number. electrons are equal to protons.

5. I don't know, sorry.

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2 years ago