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3 years ago

Complete these structures by adding electron dots as needed.

1 answer:

7 0

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

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Why is the moon visible?

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3 years ago

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Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of $4.71\times 10^{-15}J$ . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

= De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of beta particle = $9.1094\times 10^{-31} kg$

$E_k$ = kinetic energy of the particle = $4.71\times 10^{-15}J$

Putting values in above equation, we get:

$\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}$

$\lambda = 6.762\times 10^{-12} m$

$6.762\times 10^{-12} m$ is the de Broglie wavelength of this electron.

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3 years ago

PLS HELPPP why is CH3OH more polar than CH3CH2OH

dusya [7]

Answer:

Explanation:

Polarity is about differencens in electronegativity. CH bonds have around the same electronegativity value so a CH bond is nonpolar. The more CH bonds there are in a molecule, the more nonpolar it is. Since CH3CH2OH has more carbon-hydrogen bonds than CH3OH, it is more nonpolar. With the same reasoning, since CH3OH has less CH bonds, it's more polar.

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3 years ago

Iodine-131 has a half-life of 8 days. If the amount of iodine-131 in the orginial sample is 8 g, how much iodine-131 will remain

hram777 [196]

1/32

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3 years ago

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

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