Ba2Cl
NaS2
The numbers are in subscript
Answer:
A = Metallic Bond
B = Strong bonding, strong conductor, high melting and boiling points
Explanation:
Since the bond is between two metals (located in groups 11 and 12), they would experience metallic bonding. Metallically bonded molecules have high melting and boiling points due to the strength of the metallic bond. They also experience strong electrical current due to the there delocalized electrons.
Answer:
The correct answer is 169.56 g/mol.
Explanation:
Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,
= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)
= 0.0115 mole of electron
The half cell reaction for the metal X is,
X^3+ (aq) + 3e- = X (s)
From the reaction, it came out that 3 faraday will reduce one mole of X^3+.
The molar mass of X will be,
= 0.650 g/0.0115 *3 mol electron/1 mol
= 56.52 * 3
= 169.56 g/mol
Answer:
The months became darker colored and more darker colored moths were sighted.
Explanation:
All of the new air pollution caused the moths to alter there wings and bodies. The moths had no adaptation to this air pollution, causing them to be discolored.
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
