O2 is the limiting reactant
Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
Answer:
3.50 molal
Explanation:
Molality → Moles of solute / kg of solvent.
Let's convert the solvent's mass from g to kg
16.2 g . 1kg / 1000 g = 0.0162 kg
Let's determine the moles from the solute
2.61 g . 1 mol / 46 g = 0.0567 moles
Molality → 0.0567 mol / 0.0162 kg = 3.50 m
I believe the answer is a because 2 times 12 is 24. Hope this helped
