For the following reaction, 6.49 grams of potassium hydroxide are mixed with excess phosphoric acid. The reaction yields 6.86 gr
ams of potassium phosphate.
potassium hydroxide (aq) + phosphoric acid (aq) potassium phosphate (aq) + water (1)
grams
What is the theoretical yield of potassium phosphate?
What is the percent yield of potassium phosphate?
%
potassium hydroxide (aq) + phosphoric acid (aq) potassium phosphate (aq) + water (1)
grams
What is the theoretical yield of potassium phosphate?
What is the percent yield of potassium phosphate?
%
1 answer:
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Answer : Part a) Fraction of energy lost : 20 %
Part b) Speed before and after bounce = 6.3 m/s and 5.6 m/s
Part c) Energy is lost as thermal energy .
Part A) Fraction of energy lost during bouncing :
The energy possessed by any object when present at any height is potential energy . The formula of potential energy is given as :
PE = mgh
where PE = potential energy
,m = mass pf object , g = gravitational acceleration and h = height
Given : Initial height , h₁ = 2 m final height , h₂ = 1.6 m
Initial potential energy : m * g* h ₁
Final potential energy = m* g* h₂
Energy lost = Initial PE - Final PE
= ( mgh₁ - mgh2 )
Fraction of energy lost :
Plugging value in above formula and taking " mg " common =>
Fraction of energy lost =
Fraction of energy lost = 20%
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Part B ) Speed of ball just before and after the bounce.
Speed of ball before the bounce :
The potential energy gets converted to kinetic energy when it fall from height of 2m , so
Potential energy = kinetic energy
mgh₁ = m v²
or v ² = 2gh₁
Given : g = 9.8 m/s² h= 2 m
v² = 2 * 9.8 m/s² * 2 m = 39.2 m²/s²
v = 6.3 m/s
Speed of ball after bounce :
Potential energy = kinetic energy
mgh₂ = m v²
or v² = 2gh₂
= 2 * 9.8 m/s² * 1.6 m = 31.36 m²/s²
v = 5.6 m/s
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Part C) The energy lost due to friction. When the ball touches the ground , there occur friction force between the surface of ground and ball , due to which energy is lost as thermal energy .