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madam [21]
3 years ago
15

Why do the isotopes of a particular element behave differently in nuclear reactions but the same in

Chemistry
2 answers:
Sphinxa [80]3 years ago
7 0

Answer:

A and C

Explanation:

AfilCa [17]3 years ago
4 0

The isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions because the nuclear reactions involve changes to protons and neutrons in an atom and isotopes differ in neutrons and so they react differently to nuclear changes. While for chemical reactions, the electrons in an atom are involved but the isotopes do not differ in electrons and so they react the same way during chemical changes.

Answer: Option A & C

<u>Explanation:</u>

The reactants of chemical and nuclear reactions are different. The reactants of chemical reactions are the electrons present in the outermost shell of the reactants, while the reactants of nuclear reactions are either the nuclei of reactants or a nuclei and any subatomic particles of other reactants.

Thus it can be understood that chemical reactions consider the electrons while nuclear reactions consider the nuclei or the number of protons and neutrons of the reactants.

The isotopes of elements contain different mass number or we can say different number of neutrons but the number of electrons are same, so they behave differently in nuclear reactions and similar in chemical reactions.

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Which statement would support a merit of the Bronsted-Lowry base theory has over the Arrhenius base theory?
butalik [34]

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Bronsted Base is an H+ acceptor

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You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution
e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

 (10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration.  It is defined as moles/liter.  A 1 M  solution contains 1 mole of solute per liter of solvent.  [200 ml water = 0.2 Liters water.]

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8 0
1 year ago
The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
Dimas [21]
The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
6 0
3 years ago
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