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2 years ago

Consider the reaction below.

Chemistry

1 answer:

patriot [66]2 years ago

6 0

Answer:

HI and I-

Explanation:

Hello there!

In this case, according to the acid ionization of hydroiodic acid:

$HI+H_2O\rightarrow I^-+H_3O^+$

Now, since the acid, HI, results in the conjugate base, I- and the base, H2O in the conjugate acid, H3O+, it is correct to assert that the acid-conjugate base pair is HI and I-

Best regards!

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What type of reaction is CuO + H2SO4 ➡️ CuSO4 + H2O

Afina-wow [57]

Water is the answer hope this helps

7 0

3 years ago

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

Answer: The volume of hydrogen chloride produced in the reaction will be 4.4 mL

Explanation:

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

$CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)$

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

where,

$V_1\text{ and }n_1$ are the volume and number of moles of methane gas

$V_2\text{ and }n_2$ are the volume and number of moles of hydrogen chloride

We are given:

$V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol$

Putting values in above equation, we get:

$\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL$

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

5 0

2 years ago

DIA [1.3K]

V_1=49.8mL
T_1=18°C=291K
T_2=83°C=356K

Using Charles law

$\\ \sf\longmapsto V_1T_2=V_2T_1$

$\\ \sf\longmapsto V_2=V_1T_2\div T_1$

$\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}$

$\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}$

$\\ \sf\longmapsto V_2=60.9mL$

5 0

2 years ago

In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al

tekilochka [14]

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)$

here underlined is $HCH_3CO_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

Here underlined is $NH_3$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

Here underlined is $C_2H_5NH_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)$

As $HCH_3CO_2(aq)$ is donating a proton , it acts as a bronsted acid.

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

As $NH_3$ contains a lone pair of electron on nitrogen , it can easily donate electrons to $BH_3$ and act as lewi base.

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

As $C_2H_5NH_2(aq)$ is accepting a proton , it acts as a bronsted base.

7 0

2 years ago

Draw a structural formula for the alkene you would use to prepare the alcohol shown by hydroboration/oxidation.

Nastasia [14]

Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).

In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

5 0

2 years ago