You are currently viewing Lesson 22. Hydrogen Halide.  Halide Salt Pages 68, 69, 70 SBT Chemistry 10 Connecting Knowledge to Life>

Lesson 22. Hydrogen Halide. Halide Salt Pages 68, 69, 70 SBT Chemistry 10 Connecting Knowledge to Life>

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22.1 . Recognition

In the liquid state, which of the following hydrogen halide molecules can form strong hydrogen bonds?

A. HCl.

B. HI.

C. HF.

D. HBr.

Solution method:

Based on the definition of hydrogen bond: A hydrogen bond is a type of weak bond, formed between an H atom (which has bonded to a highly electronegative atom, usually F, O, N) and another atom. The other atom has a high electronegativity (usually F, O, N) and the valence electron pair has not participated in the bond

Detailed explanation:

– Answer:

Recognizing 22.2

Which of the following hydrogen halide has the highest boiling point at normal pressure?

A. HCl.

B. HBr.

C. HF.

D. HI.

Solution method:

Based on which molecule has hydrogen bond -> Higher boiling point and melting point.

Detailed explanation:

– Answer:

Recognizing 22.3

In the hydrogen halide series, from HF to HI, how does the bond strength vary?

A. Ascending.

B. Descending.

C. No change.

D. Circulatory.

Solution method:

Based on the atomic radius of the halogen group increases from fluorine to iodine → the distance between the H atom and the halogen atom increases → the H atom has increasing mobility → the bond strength decreases

Detailed explanation:

– Answer: GET

Recognizing 23.4

Which of the following solutions of hydrohalic acid is weakly acidic?

AHF.

B. HBr.

C. HCl.

D. HI.

Solution method:

Based on the atomic radius of the halogen group increases from fluorine to iodine → the distance between the H atom and the halogen atom increases → the H atom has increasing mobility → the stronger the acidity

Detailed explanation:

– Answer: A

Recognizing 22.5

Add a few drops of which of the following solutions to AgNO . solution?3 light yellow precipitate was obtained?

A. HCl.

B. NaBr.

C. NaCl.

D. HF.

Solution method:

Based on precipitation color of want silver with halide ion:

– AgCl: white precipitate

– AgBr: light yellow precipitate

– AgI: yellow precipitate

Detailed explanation:

– Answer: GET

Recognizing 22.6

In the absence of air, the iron nail reacts with HCl solution to yield the products

A. FeCl3 and THEY2.

B. FeCl2 and Cl2.

C. FeCl3 and Cl2.

D. FeCl2 and THEY2.

Detailed explanation:

– Answer: EASY

– Specifically: Fe + 2HCl → FeCl2 + FRIENDS2

Application 22.7

Hydrohalic acid is commonly used to clean metal surfaces before electroplating

A. HBr.

B. HF.

SPEND.

D. HCl.

Detailed explanation:

– Answer: EASY

Recognizing 22.8

Hydrohalic acid is used as a raw material for the production of teflon non-stick compounds

AHF.

B. HCl.

C. HBr.

D. HI.

Solution method:

Based on the formula of teflon compound is -(CF2-CF2)-n to predict

Detailed explanation:

– Answer: A

Recognizing 22.9

Which of the following solutions can distinguish F . ions?ClBrI in salt solution?

A. NaOH.

B. HCl.

C. AgNO3.

D. KNO3.

Solution method:

Distinguish F . ionsClBrI ­­by adding AgNO . solution3 into their salt solution

Detailed explanation:

– Answer: GET

Halide ion

Reagents

F

Cl

Br

I

AgNO . solution3

No phenomenon

There is a white precipitate (AgCl)

There is a light yellow precipitate (AgBr)

There is a yellow precipitate (AgI)

Get to know 22.10

KBr exhibits reducing property when heated with which of the following solutions?

A. AgNO3.

B. FAMILY2SO4 special.

C. HCl.

D. FAMILY2SO4 washy.

Solution method:

An oxidizing agent is required for reducing activity.

Detailed explanation:

– Answer: GET

Understanding 22.11

In the hydrogen halide series, from HCl to HI, the boiling point increases mainly due to which of the following?

A. Incremental van der Waals interactions.

B. Ascending Molecular Mass

C. Bond strength decreases gradually.

D. The bond polarity decreases gradually.

Solution method:

Based on

Features of van der Waals interactions:

The van der Waals interaction increases the melting and boiling points of substances

+ As the molecular weight increases, the molecular size increases, the van der Waals interactions increase

Detailed explanation:

– Answer: A

Understanding 22.12

In the hydrogen halide series, from HF to HI, how does the polarity of the bond change?

A. Cycle.

B. Ascending.

C. Descending.

D. No change.

Solution method:

Based on the electronegativity of the halogen group decreasing from fluorine to iodine → the polarity of the bond with hydrogen will decrease.

Detailed explanation:

– Answer:

Understanding 22.13

Concentrated hydrochloric acid exhibits reducing properties when reacted with which of the following substances?

A. NaHCO3.

B. CaCO3.

C. NaOH.

D. MnO2.

Solution method:

An oxidizing agent is required for reduction

Detailed explanation:

– Answer: EASY

Understanding 22.14

Dilute hydrochloric acid exhibits oxidizing properties when reacted with which of the following substances?

A. FeCO3.

B. Fe.

C. Fe(OH)2.

D. Fe2O3.

Solution method:

To exhibit oxidizing properties, a reducing agent is required.

Detailed explanation:

– Answer: GET

Understanding 22.15

Which of the following reagents can distinguish between HCl and NaCl solutions?

A. Phenolphthalein.

B. Starch paste.

C. Purple litmus.

D. Bromine water.

Detailed explanation:

– Answer:

– Explanation: HCl solution turns blue litmus red and NaCl solution does not change red litmus

Understanding 22.16

HF solution can eat glass dishes due to which of the following chemical reactions?

A. SiO2 + 4HF → SiF4 + 2 HOUSES2O.

B. NaOH + HF → NaF + H2O.

C. FAMILY2 + F2 → 2HF.

D. 2F2 + 2 HOUSES2O → 4HF + O2.

Detailed explanation:

– Answer: A

Understanding 22.17

In the hydrohalic acid series, from HF to HI, the acidity increases due to

A. ascending van der Waals interactions.

B. the bond polarity decreases gradually.

C. increasing molecular mass.

D. bond strength decreases gradually.

Solution method:

Based on the atomic radius of the halogen group increases from fluorine to iodine → the distance between the H atom and the halogen atom increases → the H atom has increasing mobility → the stronger the acidity

Detailed explanation:

– Answer: EASY

Understanding 22.18

Which of the following halide salts reacts with H . solution?2SO4 If it is dense and hot, only an exchange reaction occurs?

A. KBr.

B. KI.

C. NaCl.

D. NaBr.

Solution method:

To react with a strong oxidizing agent but only an exchange reaction occurs → the reactant has no reducing or very weak reducing properties

Detailed explanation:

– Answer:

Understanding 22.19

Which of the following statements is not correct?

A. Hydrofluoric acid solutions are corrosive to glass.

B. Solid NaCl reacts with H2SO4 concentrated, hot, hydrogen chloride is obtained.

C. Hydrogen chloride is highly soluble in water.

D. Acid force in the hydrohalic acid series decreases from HF to HI.

Detailed explanation:

– Answer: EASY

– Explanation: Acid force in the hydrohalic acid series increases from HF to HI

Understanding 22.20

Which of the following solutions can distinguish between NaF and NaCl solutions?

A. HCl.

B. HF.

C. AgNO3.

D. Br2.

Detailed explanation:

– Answer:

– Explain:

AgNO3 + NaCl → AgCl (precipitation white) + NaNO3

AgNO3 does not react with NaF because AgF is a soluble salt

Application 22.21

Carry out the hydrogen chloride solubility test according to the following steps:

– Step 1 prepare a dry flask containing HCl gas, cover the bottle with a rubber stopper with a glass tube through and a cup of water.

– Step 2: dip the glass tube into the cup of water, see the water spray into the bottle (see the picture below).

a) The phenomenon of water spraying into the tank shows that the pressure of HCl gas in the tank has increased or decreased very quickly. Explain

b) What property of HCl gas does this change in pressure demonstrate?

Detailed explanation:

a) The phenomenon of water spraying into the tank shows that the gas pressure in the tank has decreased very quickly

b) A rapid decrease in pressure indicates that hydrogen chloride gas has rapidly dissolved in water

Application 22.22

In the human body, gastric juice has an acidic environment (HCl), pH = 1.6 ÷ 2.4 to help support digestion.

a) A patient with stomach upset due to excess acid was prescribed oral medication containing NaHCO3. Write a reaction illustrating the effect of the drug.

b) At 37 °C, starch is hydrolyzed to glucose in acidic medium (HCl) with enzyme catalysis. Write the chemical equation for the reaction taking place.

Detailed explanation:

Application 22.23

There are two test tubes, each containing 2 mL saline solution of sodium Add a few drops of AgNO . solution3 into the first tube, a light yellow precipitate was obtained. Add a few drops of Cl . water2 into the second tube, shake gently, add 1 mL of benzene and shake well, the benzene changes from colorless to orange. Determine the formula of the sodium salt and write the chemical equations for the reactions taking place.

Solution method:

Based on

– Precipitation color of want silver with halide ion:

+ AgCl: white precipitate

+ AgBr: light yellow precipitate

+ AgI: yellow precipitate

– Properties of halogens: strong halogens repel weak halogens from salts

Detailed explanation:

Add a few drops of AgNO . solution3 into the first tube, a light yellow precipitate was obtained

→ The salt of sodium is NaBr

NaBr + AgNO3 → NaNO3 + AgBr (light yellow precipitate)

– Benzene from colorless to orange is because when Br2 soluble in benzene appears orange

2NaBr + Cl2 → 2NaCl + Br2

Application 22.24

Given solutions of hydrochloric acid, sodium chloride, and iodine, randomly denoted X, Y, Z.

Some experimental results are recorded in the following table.

Reagent

Reagents

Phenomena

X

starch

Appears blue purple

Z

Baking soda, NaHCO3

There are air bubbles coming out

The initial solutions are denoted by , respectively

A.Z, Y, X.

B.Y, X, Z.

C. Y, Z, X.

D. X, Z, Y.

Solution method:

– X makes starch paste turn blue-violet → X is iodine

– Z reacts with NaHCO3 foaming → Z is hydrochloric acid

NaHCO3 + HCl → NaCl + CO2 + FRIENDS2O

– And the remaining Y is sodium chlorine

→ Answer: A

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