The mass of hydrogen in 57.010 g ammonium hydrogen phosphate, (NH₄)H₂PO₄ is 2.97 g
<h3>Determination of mass of 1 mole of (NH₄)H₂PO₄ </h3>
1 mole of (NH₄)H₂PO₄ = 14 + (4×1) + (2×1) + 31 + (16×4) = 115 g
<h3>Determination of the mass of H in 1 mole of (NH₄)H₂PO₄ </h3>
Mass of H = 6H = 6 × 1 = 6 g
<h3>Determination of the mass of H in 57.010 g of (NH₄)H₂PO₄ </h3>
115 g of (NH₄)H₂PO₄ contains 6 g of H.
Therefore,
57.010 g of (NH₄)H₂PO₄ will contain = (57.010 × 6) / 115 = 2.97 g of H
Thus, 2.97 g of Hydrogen, H is present in 57.010 g of (NH₄)H₂PO₄
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Answer:
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Answer:
B. databases of periodicals
Explanation:
Databases include the citations of articles in magazines, journals and newspapers. A periodical is a magazine or newspaper published at regular intervals. So if she's looking for news articles about this earthquake then she should most likely search databases or periodicals.
If A ∆ B = A × (A + 2B) = A² + 2AB, then
• 1 ∆ B = 1 + 2B
• 4 ∆ (1 ∆ B) = 4 ∆ (1 + 2B) = 24 + 16B
• 2 ∆ (4 ∆ (1 ∆ B)) = 2 ∆ (24 + 16B) = 100 + 64B
