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klemol [59]
3 years ago
8

Identify h2so4(aq) as an acid or a base. write a chemical equation showing how this is an acid according to the arrhenius defini

tion.
Chemistry
2 answers:
Norma-Jean [14]3 years ago
5 0

{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is an \boxed{{\text{acid}}}. The chemical equation that shows the acidic character of {{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} in accordance with Arrhenius definition is \boxed{{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{{\text{H}}^ + } + {\text{SO}}_4^{2 - }}.

Further explanation:

Arrhenius theory:

According to this theory, acid produces hydrogen or hydronium ions in a solution, while the base produces hydroxide ions in a solution. HCl, and {\text{HN}}{{\text{O}}_3} are Arrhenius acids while NaOH and KOH are the examples of Arrhenius bases.

Consider the example of HCl. Its dissociation occurs as follows:

{\text{HCl}}\to{{\text{H}}^+}+{\text{C}}{{\text{l}}^-}

HCl releases {{\text{H}}^+} ions on dissociation and is therefore an Arrhenius acid.

The dissociation of NaOH occurs as follows:

{\text{NaOH}}\to{\text{N}}{{\text{a}}^+}+{\text{O}}{{\text{H}}^-}

NaOH dissociates to produce hydroxide ions and is therefore an Arrhenius base.

The dissociation of {{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} occurs as follows:

{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{{\text{H}}^+}+{\text{SO}}_4^{2-}

One mole of {{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} dissociates to produce two moles of hydrogen ions \left({{{\text{H}}^+}}\right) and one mole of sulfate ions \left({{\text{SO}}_4^{2-}}\right) According to Arrhenius theory, any species that produces hydrogen ions in solution is termed as an acid. Therefore {{\mathbf{H}}_{\mathbf{2}}}{\mathbf{S}}{{\mathbf{O}}_{\mathbf{4}}} is an acid.

Learn more:

1. The reason for the acidity of water brainly.com/question/1550328

2. Reason for the acidic and basic nature of amino acid. brainly.com/question/5050077

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acid, base and salts

Keywords: Arrhenius theory, chemical equation, H2SO4, H+, OH-, hydrogen, hydroxide, HCl, NaOH, HNO3, KOH, Cl-, dissociation, Arrhenius acid, Arrhenius base, hydronium ions, one mole, two moles.

yarga [219]3 years ago
4 0
  H2SO4  is an  acid

the  chemical  equation  showing  how H2SO4  is an acid  according  to the Arrhenius  definition   is as below

H2SO4  dissociate to  give  2H^+  and  SO4^2-

that is   H2SO4  = 2H^+  +  SO4^2-

According to Arrhenius  an acid  dissociate  to give H^+  ions H2SO4 is an acid  since  it dissociate  to  give  two hydrogen ions  
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3 years ago
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

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This is because in the graph shown line A has a quite greater impact of refraction than line B .

Hence, we can conclude that line A has the greater reaction at a faster rate.

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