Answer:
The coefficient of friction of the surface μ = 0.235
Explanation:
Values of the given variables:
Weight of block = 8.5 N
Pushing force F = 2 N
velocity = constant
Solution:
Newton's Second Law of motion states that acceleration of an object depends on the mass of the object and the force applied
F = m a
Since v =constant, a = 0. Thus
Net force Fnet = 0.
Now F_net is the sum of the pushing force and the frictional force along the horizontal, which opposes the motion:
Fnet = Fpushing - Ffrictional_force
Given that Fnet = 0, Fpushing = Ffrictional_force. and Ffrictional_force = 2 N
Equation of friction is given by
Ffrictional_force = μ*W
2 = μ*8.5
μ = 2/8.5 = 0.235
Answer:
Explanation:
2 NO2(g) ⇄ N2O4(g)
Adding Argon to this reaction will have NO effect. Catalysts nor inert gases have an affect on equilibrium conditions.
Only changes in concentration, temperature conditions and pressure-volume conditions (unless both sides have equal molar volumes) will affect the equilibria.
NH4OH(aq) ⇄ NH3(g) + H2O(l)
Removing ammonia from reaction equilibrium causes the reaction to shift right to replace removed ammonia. => Think of the reaction as being on a seesaw => removing ammonia from the product side tilts the seesaw left causing the NH₄OH to decompose and deliver more NH₃ and H₂O to the product side to increase weight on that side and level the seesaw. :-)
Energy and oreantation its one if not bolth of those
Answer:
6.8 × 10⁻⁴ mol/L.atm
Explanation:
Step 1: Given data
Solubility of nitrogen gas at 25°C (S): 4.7 × 10⁻⁴ mol/L
Partial pressure of nitrogen gas (P): 522 mmHg
Step 2: Convert the partial pressure of nitrogen to atm
We will use the relationship 1 atm = 760 mmHg.
Step 3: Calculate the value of the Henry's Law constant (k)
We will use Henry's law.
