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pav-90 [236]
3 years ago
13

Which compounds are the Bronsted- Lowry bases in the equilibrium?

Chemistry
2 answers:
elena-s [515]3 years ago
6 0

Answer : Option D) H_{2}BO_{3}^{-} and C_{2}O_{4}^{2-}.

Explanation : The acid base theory of Bronsted-Lowry, stated the definition of acids and bases on the basis of conjugated.

The fundamental concept of this theory states that when an acid and a base reacts with each other, the acid forms its conjugate base, and the base forms its conjugate acid by exchanging the proton.

So, in the given reaction;

HC_{2}O_{4}^{-} + H_{2}BO_{3}^{-}  H_{3}BO_{3} + C_{2}O_{4}^{2-}

It is firstly observed that the species H_{2}BO_{3}^{-} is accepting a proton from the acid, therefore this is one of the bases.

Secondly,C_{2}O_{4}^{2-} accepts the proton and forms the reactant again in the reversible reaction.

Therefore, these are the correct options.

H_{2}BO_{3}^{-} and C_{2}O_{4}^{2-}

Butoxors [25]3 years ago
4 0
Its D for plato. 
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Which best describes a difference between energy transformations in power plants and dams? Only power plants use fossil fuels to transform energy. Only dams use fission to generate thermal energy. ... Only dams use mechanical energy to produce electricity.

Explanation:

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3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

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3 years ago
nuclear reactions can be written out in equation form with information about the nuclei that are produced. in one to two sentenc
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Q = 100 g • 0,46 J/g•°C • 25,2°C
Q = 1160 J.
C - specific heat.

3 0
2 years ago
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