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DaniilM [7]
2 years ago
9

The individual dipole moments in ammonia (NH3) do not cancel each other

Chemistry
1 answer:
kozerog [31]2 years ago
4 0

Answer:

                    The strongest force that exists between molecules of Ammonia is <em>Hydrogen Bonding</em>.

Explanation:

                    Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atoms (i.e. F, O and N) of one molecule interacts with the partial negative most electronegative atom of another molecule.

                    Hence, in ammonia the nitrogen atom being more electronegative element than Hydrogen will be having partial negative charge and making the hydrogen atom partial positive. Therefore, the attraction between these partials charges will be the main force of interaction between ammonia molecules.

                  Other than Hydrogen bonding interactions ammonia will also experience dipole-dipole attraction and London dispersion forces.

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A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is
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Answer #1 is "there is 2.5 grams of solute in every 100 g of solution." 
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution." 
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
     mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
                                                          = 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
     mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
                                                          = 25g solute/(25g solute + 975g solvent)
                                                          = 25g solute/1000g solution
                                                          = 25g solute/1kg solution
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4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
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464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

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