Answer: By carrying out series of wet test
Explanation: Dissolve the ionic compound in water. Ionic compounds will ionise and allow the solution to conduct an electric current. Afterwards specific test will be carried out to detect the presence of cations and anions. Another way would be to find the substance's melting point. Ionically bonded compounds have much higher melting points.
Answer:
1. Answer: The bowling ball has more potential energy as it sits on top of the building. It does not have any kinetic energy because it is not moving.
2. Answer: The bowling ball has equal amounts of potential and kinetic energy half way through the fall. At the half way point, half of the potential energy has been converted to kinetic energy.
3. Answer: Just before the ball hits the ground, it has more kinetic energy. As it hits the ground the potential energy becomes zero.
4. Answer:
PE=784 J
5. Answer:
PE = 392 J
6. Answer:
KE= 392 J
Also, since the PE and KE are equal at the half way point and PE =392 J, KE = 392 J.
7. What is the kinetic energy of the ball just before it hits the ground?
Answer:
KE=784 J
At first I answered in the comments, but I am able to answer now. I hope this can help
Answer:
Anode half reaction equation:
Ni(s)------> Ni^2+(aq) + 2e-
Explanation:
Looking at the values of reduction potential given in the question, Ni2+|Ni half cell has a more negative reduction potential than the Fe3+|Fe2+ half cell. The more negative the reduction potential of a half cell, the more its tendency to act as the anode. Hence based on the half cell reduction potentials presented in the question, Ni2+|Ni is the anode while the other is the cathode.
Answer:
4.21 g of AgCl
3.06 g of BaCl₂ will be needed to complete the reaction
Explanation:
The first step is to determine the reaction.
Reactants: BaCl₂ and AgNO₃
The products will be the silver chloride (AgCl) and the Ba(NO₃)₂
The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)
We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-
2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.
We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.
Now, we consider the BaCl₂.
2 moles of nitrate can react to 1 mol of barium chloride
Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:
0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂
