Answer:
final concentration: Ca = 0.014 M
Explanation:
Velocity of reaction:
∴ α: order of reaction, assuming α = 1
∴ K = 0.249 s-1.......rate constant
∴ Cao = 0.050 M......initial concentration
∴ t = 5 s.......reaction time
⇒ δCa/δt = K*Ca
⇒ ∫δCa/Ca = K*∫δt
⇒ Ln(Cao/Ca) = K*t = (0.249s-1)(5 s) = 1.245
⇒ Cao/Ca = 3.473
⇒ Ca = 0.050/3.473
⇒ Ca = 0.014 M
There are approximately 160 grams in 1 mol of Fe2O3 molecules. Therefore, there would be 79/160= 0.49375 mols of Fe2O3 molecules in 79 grams. There are 5 atoms in total for each molecule of Fe2O3, therefore 79/160 * 5 = 79/32 = 2.46875 mols of atoms.
Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:
Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
of HCL
According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
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
