All categories

3 years ago

Write a net ionic equation to show that hydroiodic acid, hi, behaves as an acid in water.

Chemistry

2 answers:

Marianna [84]3 years ago

7 0

The answer is:

HI (aq) + H₂O (ℓ) ⟶ H₃O⁺ (aq) + I⁻ (aq)

<h2>Further Explanation </h2>

In chemistry, the equation of a reaction or chemical equation is the symbolic writing of a chemical reaction. The chemical formula of the reagent is written to the left of the equation and the chemical formula of the product is written to the right. The coefficient written to the left of a chemical formula is a stoichiometric coefficient, which describes the amount of that substance involved in a reaction relative to another substance. The reaction equation was first made by the iatrochemist Jean Beguin in 1615. In a reaction equation, reagents and products are connected by different symbols. The symbol → is used for one-way reactions, ⇆ for two-way reactions, and ⇌ equilibrium reactions.

Clean ionic equations are important aspects for chemistry because these equations only represent the states of matter that change in chemical reactions. This equation is commonly used in redox reactions, double replacement reactions, and acid-base neutralization. There are three basic steps to writing a clean ionic equation: balancing a molecular equation, converting it into a fully ionic equation (how each type of substance is in solution), and Determining the cations and anions in a compound.

1. equalizing molecular equations

The first step to writing a clean ionic equation is to identify the ionic compounds from the reaction. Ionic compounds are compounds that will ionize in aqueous solution and have a charge. Molecular compounds are compounds that have never had a charge. These compounds are formed from two nonmetals and are often referred to as covalent compounds.

Ionic compounds can be formed from metals and non-metals, metal and polyatomic ions, or some polyatomic ions.

If you are not sure about a compound, look for elements of the compound in the periodic table.

2. transform into a fully ionic equation

Not all ionic compounds can dissolve in aqueous solutions. Thus, the compound will not dissolve into individual ions. You must identify the solubility of each compound before continuing the rest of the equation. The following is a summary of the rules regarding solubility.

3. Determine the cations and anions in a compound

Cations are positive ions in a compound and are usually metals. Anions are nonmetal negative ions in compounds. Some non-metals can form cations, but metals will always form cations.

Learn more

definition of chemical equation brainly.com/question/11575465

three basic steps to writing a clean ionic equation brainly.com/question/11575465

Details

Grade: College

Subject: Chemistry

keywords: chemical equation

OlgaM077 [116]3 years ago

3 0

Answer:

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Explanation:

The HI donates a proton to the water, converting it to a hydronium ion

HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)

Thus, the HI is behaving like a Brønsted acid.

You might be interested in

Which functional group is found in an ester?

Klio2033 [76]

<h2>QUESTION:- Which functional group is found in an ester?</h2>

ESTER IS THE COMPOUND IN WHICH OH GROUP IS REPLACED BY THE (O) in the substitution reaction.

so common formula of ester ->

RCOOR'

IN WHICH R AND R' ARE CARBON CHAINS

4 0

2 years ago

Read 2 more answers

What is the usual charge on an ion from group 7a

alexdok [17]

Group 7a would have an ion charge of -1 because it has 7 valence electrons and it wants to gain one more electron(which is negative) to have a full shell of 8

4 0

3 years ago

1) Какое количество вещества Феррум (III) гидроксида вступает в реакцию с 15 г сульфатной кислоты?

EleoNora [17]

i dont speak russian sorry

3 0

3 years ago

4) What mass of hydrogen is formed when 2.00g of magnesium<br> Mg + H2SO4–>H2SO4

DochEvi [55]

Answer:

0.16g

Explanation:

Please mark as brainliest

7 0

2 years ago

Read 2 more answers

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago