Answer:
C: The temperature of the substance increases as it sits in the beaker of water
Explanation:
This question was taken from a video where an attempt was made to investigate the changes in temperature when a substance undergoes change from it's solid phase to its liquid phase.
To do this, as seen in the video online, it shows a solid substance in a test tube being placed in a beaker of water.
From observation, the water in the beaker has a warmer temperature than the solid substance present in the test tube and this in turn makes the test tube gradually increase in temperature.
Thus, the solid substance will as well increase increase in temperature when it is placed in the beaker of water.
Answer:
- The chemical reaction is not balanced. There is two oxygens on the reactant's side while there's only one oxygen on the products side.
- I would not say it's following the law of conservation of mass as it's not a balanced equation.
- To balance this equation, you would need to add the coefficient of '2' to Magnesium (Mg) on the reactants side, and add the coefficient of '2' to the products side. This would make it so that there's 2 Mg's and 2 O's on both the reactant's side and products side.
edit: I hope this helped you in some way. ^^
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34
= 0.059 g
percent morphine =
=
= 8.6 %
Answer:
7.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.
Thus, 7.5 moles of O₂ were obtained from the reaction.
