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Mkey [24]
3 years ago
10

Many free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical–radic

al combination reaction is the formation of a new electron‑pair bond. Consider the formation of hydrogen peroxide. 2OH(g)⟶H2O2(g) Write Lewis formulas for the reactant and product species in the chemical equation. Include nonbonding electrons.

Chemistry
1 answer:
Alexxandr [17]3 years ago
4 0

Answer:

In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.

Explanation:

Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.

The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.

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Need help with chemistry hw please help
Andru [333]
Co2 = two covalent bonds
ccl4 = 4 covalent bonds
Lih = covalent bond
5 0
3 years ago
Read 2 more answers
Calculate the mass in grams of each sample.<br> 4.88x10^20 H2O2 molecules
AleksAgata [21]

4.88x10^20 H2O2 molecules

5 0
3 years ago
6. An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. To p
pishuonlain [190]

Answer:

[Na₂CO₃] = 0.094M

Explanation:

Based on the reaction:

HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)

It is possible to find pH using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base,  [CO₃²⁻] = [Na₂CO₃] and  [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.

pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>

<em />

Replacing these values:

10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]

<em> [Na₂CO₃] = 0.094M</em>

<em />

5 0
3 years ago
antimony has two naturally occuring isotopes, sb121sb121 and sb123sb123 . sb121sb121 has an atomic mass of 120.9038 u120.9038 u
Luda [366]

Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

  • sb121 has an atomic mass of 120.9038 u.
  • sb121 has a percent natural abundance of x.
  • sb123 has an atomic mass of 122.9042 u.
  • sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
  • Antimony has an average atomic mass of 121.7601 u

The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

Learn more about average atomic mass:

brainly.com/question/4923781

brainly.com/question/1826476

brainly.com/question/15230683

brainly.com/question/7955048

#SPJ1

5 0
2 years ago
Quick electron emissions are called
Montano1993 [528]
Defined as a phenomenon of liberation of electron from the surface that is stimulated by temperature elevation, radiation, or by strong electric field.
3 0
3 years ago
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