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
Learn more about mass composition:
brainly.com/question/13531044
Answer: See below
Explanation:
For this question, we are given the volume, and we want to find the diameter. We have to work backwards to get our radius, then to find the diameter.
(Volume of cylinder)
(The π cancel out)
(Plug in h=8)
(Divide 8)
(Square root of 9 is 3)
Now that we have our radius, we can find our diameter.
(Equation for diameter)
(Plug in r=3)
(Multiply)
The answer would be 20 percent
