Answer:
A. Endothermic reaction.
B. +150KJ.
C. 250KJ.
Explanation:
A. The graph represents endothermic reaction because the heat of the product is higher than the heat of the reactant.
B. Determination of the enthalpy change, ΔH for the reaction. This can be obtained as follow:
Heat of reactant (Hr) = 50KJ
Heat of product (Hp) = 200KJ
Enthalphy change (ΔH) =..?
Enthalphy change = Heat of product – Heat of reactant.
ΔH = Hp – Hr
ΔH = 200 – 50
ΔH = +150KJ
Therefore, the enthalphy change for the reaction is +150KJ
C. The activation energy for the reaction is the energy at the peak of the diagram.
From the diagram, the activation energy is 250KJ.
Answer:
The distance traveled by the balloon is 10.77 m
Explanation:
velocity of the ball, 
= 2 m/s south
velocity of the air, 
= 5 m/s west
To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.
| 2m/s
|
|
↓
5m/s ←------------------
the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.
R² = 5² + 2²
R² = 29
R = √29
R = 5.385 m/s
The distance traveled by the balloon is calculated as;
d = R x t
where;
t is time of the motion = 2 seconds
d = 5.385 x 2
d = 10.77 m
Therefore, the distance traveled by the balloon is 10.77 m.
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
