Asked
9 months ago

Viewed
39433 times

Answer:

2. 7.39×10²⁰ atoms of iodine

3. 1.04×10¯¹⁶ g of Cadmium

4. 1.53×10²⁸ atoms of antimony.

Explanation:

2. Determination of the number atoms of iodine in 0.156 g of iodine.

From Avogadro's hypothesis,

1 mole of iodine = 6.02×10²³ atoms

1 mole of iodine = 127 g

Thus,

127 g = 6.02×10²³ atoms

Therefore,

0.156 g = 0.156 g × 6.02×10²³ atoms / 127 g

0.156 g = 7.39×10²⁰ atoms

Thus, 0.156 g of iodine contains 7.39×10²⁰ atoms

3. Determination of the mass of cadmium that contains 556000 atoms.

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole Cd

1 mole of Cd = 112.5 g

Thus,

6.02×10²³ atoms = 112.5 g

Therefore,

556000 atoms = 556000 atoms × 112.5 g / 6.02×10²³ atoms

556000 atoms = 1.04×10¯¹⁶ g

Thus, 1.04×10¯¹⁶ g of Cd contains 556000 atoms.

4. Determination of the number of atoms of antimony in 3.1×10⁶ g of antimony.

From Avogadro's hypothesis,

1 mole of antimony = 6.02×10²³ atoms

1 mole of antimony = 122 g

Thus,

122 g = 6.02×10²³ atoms

Therefore,

3.1×10⁶ g = 3.1×10⁶ g × 6.02×10²³ atoms / 122 g

3.1×10⁶ g = 1.53×10²⁸ atoms.

Thus, 3.1×10⁶ g of antimony contains 1.53×10²⁸ atoms.

PLEASE HELP

PLEASE HELP

Enter your username or email to reset your password. You will receive an email with instructions on how to reset your password. If you are experiencing problems resetting your password contact us or send us an email