Answer:
1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
Explanation:
Given data:
mass of cadmium = 6.35 g
Number of atoms of aluminum as 6.35 g cadmium contain = ?
Solution:
Number of moles of cadmium = 6.35 g/ 112.4 g/mol
Number of moles of cadmium = 0.06 mol
Number of atoms of cadmium:
1 mole = 6.022×10²³ atoms of cadmium
0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol
0.36×10²³ atoms of cadmium
Number of atoms of Al:
Number of atoms of Al = 0.36×10²³ atoms
1 mole = 6.022×10²³ atoms
0.36×10²³ atoms × 1 mol /6.022×10²³ atoms
0.06 moles
Mass of aluminum:
Number of moles = mass/molar mass
0.06 mol = m/ 27 g/mol
m = 0.06 mol ×27 g/mol
m = 1.62 g
Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole
D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles
<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>
1 mole of N₂H₄ contains 4 moles of H
Therefore,
0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H
<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>
1 mole of C₁₀H₂₂ contains 22 moles of H
Therefore,
34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H
Learn more about mole:
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Answer:
diboron trioxide Formula: B 2 O 3 Molecular weight: 69.620 CAS Registry Number: 1303-86-2
Explanation:
<h3>
<u><em>hope that helps you</em></u>╰(*°▽°*)╯</h3>
Answer:
I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.
