<u>Answer:</u>
<em>Equivalence point and end point are terminologies in pH titrations and they are not the same.
</em>
<u>Explanation:</u>
In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.
At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the <em>pH at equivalence point. </em>
<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution.
</em>
Answer:
<h3>0.445</h3>
Explanation:
In friction, the coefficient of friction formula is expressed as;
Ff is the frictional force = Wsinθ
R is the reaction = Wcosθ
Substitute inti the equation;
Given
θ = 24°
Hence the coefficient of kinetic friction between the box and the ramp is 0.445
Answer
given,
gauge pressure = 1.94 x 10⁵ Pa
Pressure due to 4.90 m column of water
= ρ g h
= (4.90) x (1000) x (9.8) Pa
= 48020 Pa
Gauge pressure of second floor faucet
= 1.94 x 10⁵Pa - 48020 Pa
P_g= 145980 Pa
( b )
Let h = height of faucet from which no water can flow even if open
P = ρ g h
1.94 x 10⁵ = h x(1000) x (9.8)
h = 19.79 m
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
