Can someone help with this 3 question Thankyou god bless
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.
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