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# Can someone help with this 3 question Thankyou god bless

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Can someone help with this 3 question Thankyou god bless

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2. 7.39×10²⁰ atoms of iodine

4. 1.53×10²⁸ atoms of antimony.

Explanation:

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

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.

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.

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.

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