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3 years ago

8

When Phosphoric Acid behaves according to arrhenius theory in water, what are the two products formed by the first proton

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

Phosphoric acid is a triprotic acid. This means that it can dissociate in water up to three times, each time releasing a proton into the water as shown in the following reactions:

H3PO4 (s) + H2O (l) is in equilibrium with H3O + (aq) + H2PO4− (aq)

H2PO4− (aq) + H2O (l) is in equilibrium with H3O + (aq) + HPO42− (aq)

HPO42− (aq) + H2O (l) is in equilibrium with H3O + (aq) + PO43− (aq)

Explanation:

Phosphoric acid having contact with water, dissociating from a proton up to three times, that is why the three possible reactions are determined above.

This acid is an acid that belongs to oxo acids and its formula is H3PO4

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Answer:

3

Explanation:

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3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

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3 years ago

A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K

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Answer:

-0.050 kJ/mol.K

Explanation:

A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K

The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

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3 years ago

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Write the chemical formula for the cation present in the aqueous solution of cuso4.

Inessa05 [86]

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So, the chemical formula of the cation present in the aqueous solution of $CuSO_4$ is $Cu^{2+}$ .

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3 years ago

What is the name of CH3CH2CHMgBr(CH3)?

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Answer:

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1 year ago