The answer is: (5696 J) / (155 g) / (40.0 - 25.0)°C = 2.45 J/g·°C
Answer:
percentage of water = 18.76%
Explanation:
A chemist has a sample of hydrated Li2SiF6 and it weighs 0.4813 grams.
The overall weight of the compound is 0.4813 grams.
weight of hydrated sample = 0.4813 grams.
weight of anhydrous compound = 0.391 grams
percentage weight of water = mass of water/mass of the hydrated compound × 100
mass of water = mass of hydrated compound - mass of anhydrous compound
mass of water = 0.4813 - 0.391 = 0.0903 grams.
percentage of water = 0.0903/0.4813 × 100
percentage of water = 9.03/0.4813
percentage of water = 18.761687097
percentage of water = 18.76%
This is known as Rutherford's gold foil experiment. To align with J.J Thompson's Plum Pudding Model, he expects a beam of alpha particles to just pass through the gold foil undisturbed. However, some were deflected at certain angles. Alpha particles are positive, so it would just go straight through the nucleus, but will deflect if it hits the electrons. <em>Therefore, the answer is: </em><span><em>Particles that struck the nucleus went straight.</em></span>
Answer:
0.190L of hydrogen may be produced by the reaction.
Explanation:
Our reaction is:
3Mg + 2H₃PO₄ → Mg₃(PO₄)₂ + 3H₂
We need to determine the limting reactant. Let's find out the moles of each:
5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg
55.23 g . 1 mol / 97.97 g = 0.563 moles of acid
2 moles of acid react to 3 moles of Mg
0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg
Definetely the limting reactant is Mg.
As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen
Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂
At STP, 1 mol of any gas occupies 22.4L
0.00857 mol . 22.4L / 1mol = 0.190L
