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raketka [301]
2 years ago
14

Which compound could serve as a reactant in a neutralization reaction ?

Chemistry
1 answer:
Veseljchak [2.6K]2 years ago
7 0
 The   compound that  could serve  as  a reactant   in the neutralization reaction is    H2SO4
     
  Explanation
   Neutralization reaction occur  between an acids  and a base. H2SO4             (  sulfuric acid)  is  a strong  acid.  It can be neutralized by  strong base such as NaOH ( sodium  hydroxide)

  Example of neutralization reaction is

2NaOH + H2SO4 → Na2SO4  + 2H2O
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We saw in the calorimeter experiment that the combustion of a fuel is exothermic. what can you then say about the formation of a
ankoles [38]

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, fuel is formed through a natural phenomenon involving the conversion of large amount dead and decayed organisms (usually algae and zooplanktons) to combustible fuel through exposure to relatively high temperature and pressure (over millions of years) in the earth's crust. Thus, since this involves a sort of absorption of heat energy (from the earth's crust), it can be referred to be an endothermic reaction.

8 0
2 years ago
Compared to the nucleus 5626fe, what is the density of the nucleus 112 48cd?
tia_tia [17]

For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.

n(Cd) / n(Fe)=1

<h3>What is the density of the nucleus 112 48cd?</h3>

Generally, the equation for the density  is mathematically given as

d=\frac{A}{4/3}\piR^3

Therefore

n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3

n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3

n(Cd) / n(Fe)=1

In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same

Read more about density

brainly.com/question/14010194

8 0
2 years ago
34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i
nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

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     \frac{dp(r)}{dr} = 0

            Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}} + Ae^{\frac{-r}{3a_{o}}}(3r^{2}) = 0

                     \frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}

                      = 3Ar^{2}e^{\frac{-r}{3a_{o}}}

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Thus, we can conclude that most likely radius at which the electron would be found is 9a_{o}.

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