The height of the ball at time t is given as h = -16t² + 14t + 6 while the height of the receiver hand at time t is given as h = -16t² + 10t + 8
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
A quarterback throws a football toward a receiver from a height of 6 ft. The initial vertical velocity of the ball is 14 ft/s. At the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of 10 ft/s.
The height of the ball at time t is given as h = -16t² + 14t + 6 while the height of the receiver hand at time t is given as h = -16t² + 10t + 8
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Considering the definition of percentage by mass, the sample of pure lithium nitrate contains 7.99% lithium by mass.
<h3>Definition of percentage by mass</h3>
The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
<h3>This case</h3>
Because percent by mass indicates the ratio of solute to solution based on mass, if the sample has twice the mass of the first sample, the ratio remains constant.
So the percentage by mass is the same.
That is, the sample of pure lithium nitrate contains 7.99% lithium by mass.
.
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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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