Decomposition of Copper(II) Chloride

Decomposition of Copper(II) Chloride

Assessment

Interactive Video

Chemistry

9th - 10th Grade

Hard

Created by

Jackson Turner

FREE Resource

The video tutorial explains how to balance the chemical equation for the decomposition of copper(II) chloride (CuCl2) into copper(I) chloride and chlorine gas. The reaction occurs at high temperatures, around 1000 degrees Celsius. The instructor demonstrates the process of counting atoms on both sides of the equation and adjusting coefficients to achieve balance, emphasizing the importance of converting odd numbers to even for easier balancing. The tutorial concludes with a balanced equation and tips for handling similar problems.

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7 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main product formed when CuCl2 decomposes?

Copper(II) nitrate

Copper(II) sulfate

Copper(I) chloride

Copper(II) oxide

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

At what temperature does the decomposition of CuCl2 occur?

1250 degrees Celsius

1000 degrees Celsius

750 degrees Celsius

500 degrees Celsius

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it challenging to balance the initial equation for CuCl2 decomposition?

There are too many copper atoms.

The chlorine atoms are in an odd number.

The reaction produces too much heat.

The equation is already balanced.

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step taken to balance the chlorine atoms in the equation?

Add more copper atoms.

Remove chlorine gas.

Double the copper(I) chloride.

Add more heat.

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How many chlorine atoms are present on the product side after balancing?

Five

Four

Three

Two

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the final step to ensure the equation is balanced?

Remove excess chlorine gas.

Add a coefficient of two in front of copper(II) chloride.

Adjust the temperature.

Add more copper(II) chloride.

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is it helpful to convert odd numbers to even when balancing equations?

It makes the equation more complex.

It simplifies the balancing process.

It decreases the reaction temperature.

It increases the reaction rate.