Chemistry, 23.10.2019 18:30 jasmin2344

When elemental boron, b, is burned in oxygen gas, the product is diboron trioxide. if the diboron trioxide is then reacted with a measured quantity of water, it reacts with the water to form what is commonly known as boric acid, b(oh)3. write a balanced chemical equation for each of these processes. (use the lowest possible coefficients. omit states-of-matter in your answer.)

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Answer from: jacobrobles755

The chemical equation is written below.

Explanation:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.

The chemical equation for the reaction of elemental boron and oxygen gas follows:

$4B+3O_2\rightarrow 2B_2O_3$

By Stoichiometry of the reaction:

4 moles of elemental boron reacts with 3 moles of oxygen gas to produce 2 moles of diboron trioxide.

The chemical equation for the reaction of diboron trioxide and water follows:

$B_2O_3+3H_2O\rightarrow 2H_3BO_3$

By Stoichiometry of the reaction:

1 mole of diboron trixoide reacts with 3 moles of water to produce 2 moles of boric acid.

Hence, the chemical equations are written above.

Answer from: Quest

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

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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When elemental boron, b, is burned in oxygen gas, the product is diboron trioxide. if the diboron tr...

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