Answer:
As the amplitude of pendulum motion increases, the period lengthens, because the restoring force −mgsinθ increases more slowly than −mgθ (sinθ≅θ−θ3/3!for small angles).
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
Answer:
see explanation
Explanation:
The reaction has a negative rate law; i.e., Rate = - ΔConcentration / ΔTime which is graphically a negative slope for the plot of Rate as a function of reactant concentration. => Rate ∝ f(Reactant Concentration). However, by raising the temperature, an increase the probability of reaction occurs and the formation of more product.
Answer:
43.5 moles of HNO₃.
Explanation:
The balanced equation for the reaction is given below:
S + 6HNO₃ —> H₂SO₄ + 6NO₂ + 2H₂O
From the balanced equation above,
6 moles of HNO₃ reacted to produce 2 moles of H₂O.
Finally, we shall determine the number of mole of HNO₃ required to produce 14.5 moles of H₂O.
This can be obtained as illustrated below:
From the balanced equation above,
6 moles of HNO₃ reacted to produce 2 moles of H₂O.
Therefore, Xmol of HNO₃ will react to produce 14.5 moles of H₂O i.e
Xmol of HNO₃ = (6 × 14.5)/2
Xmol of HNO₃ = 43.5 moles
Therefore, 43.5 moles of HNO₃ is required to produce 14.5 moles of H₂O.
