Chemistry, 28.07.2020 21:01 annafran89

The lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group (red). in this representation, each y atom needs 1 electron(s) to complete its octet, and gains these electrons by forming 1 bond(s) with atoms of h . there are 1 unshared electron pair(s) and 1 bonding electron pair(s) in the product molecule. the bonds in the product are .

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The molecule has been $\rm H_2Y$ , where the two bonds have been formed for the sharing of 2 electrons pair between the H and Y atom.

The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group ______ (red).

The main group element that has the ability to form $\rm H_2Y$ type of structure has been Group 16.

In this representation, each Y atom needs ______ electron(s) to complete its octet, and gains these electrons by forming______ bond(s) with atoms of H .

The atom required 2 electrons to complete its octet, and in order to gain two electrons, the atom form 2 bonds with the H, resulting in $\rm H_2Y$ .

There are ______ unshared electron pair(s) and _______ bonding electron pair(s) in the product molecule.

The atom forms bond with the two H atoms, thus there has been the presence of two unshared electrons pair. The bonding electrons pair are also two in the molecule.

The bonds in the product are _________ (Ionic or Covalent)

The bond has been formed with the sharing of the electron pair. Thus it has been a covalent bond.

For more information about the bonds in the molecule, refer to the link:

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Answer from: daniellaZemira

The question is incomplete, the complete question is;

The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group (red).

In this representation, each Y atom needs electron(s) to complete its octet, and gains these electrons by forming bond(s) with atoms of H .

There are unshared electron pair(s) and bonding electron pair(s) in the product molecule.

The bonds in the product are (Ionic or Covalent)

1) 16

2) 2 electrons

3) 2 bonds

4) 2 unshared pairs of electrons

5) 2 bonding pairs of electrons

6) The bonds in the product are covalent

Explanation:

Group sixteen elements have six electrons on their outermost shell. These include two unshared pairs of electrons and two unpaired electrons. These two unpaired electrons can now be covalently bonded to two hydrogen atoms to give H2Y. The compound H2Y has two lone pairs and two bond pairs of electrons.

H2Y can be a general formula for all hydrides of group 16. They are all very similar in structure but gradually differ in physical and chemical properties according to the graduated variation observed down the group.

Answer from: Quest

Mass is the amount of matter in an object. weight is the measurement of the pull of gravity on an object. say for example, you travel to the moon which has a different gravitational field compared to the gravitational field of earth. (gravitational fields: earth = 1, moon = 0.165) now when you are on the moon, the pull of gravity on you is different than on the earth. and as weight is the measurement of the pull of gravity on an object, that means yes, your weight can and will change on the moon. however, on the moon, on earth or even on jupiter, the amount of mass inside of you remains the same. meaning your physical mass remains constant. in conclusion it is possible for your weight to change as your mass remains constant.

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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The lewis representation above depicts a reaction between hydrogen (blue) and a main-group element f...

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