Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
Since there is more energy added as heat rises, the particles disperse and have larger movements.
This follows the law of conservation of momentum. Momentum is the product of mass and velocity of object.
Momentum = mass(m) x velocity(v)
law of conservation of momentum means that the total momentum of system before the collision of 2 objects is equal to the total momentum after the collision
Before the collision total momentum
= m1v1 + m2v2
m1 = 2 kg
v1 = 2 m/s
m2 = 6 kg
v2 = 0 m/s
substituting the values in the equation
total momentum before = (2 kg x 2 m/s) + (6 kg x 0 m/s)
total momentum = 4 kgm/s
after the collision the 2 objects stick together and have a common velocity
total momentum after the collision = (6 kg + 2 kg)x V = 8V
V = speed of the conglomerate particle
since total momentum before is equal to total momentum after
8V = 4
V = 2 m/s
speed of conglomerate particle is 2 m/s
