Answer:
I A hole
Explanation:
I must be in a museum, because you truly are a work of art
Answer:
The molarity of the dissolved NaCl is 6.93 M
Explanation:
Step 1: Data given
Mass of NaCl = 100.0 grams
Volume of water = 100.0 mL = 0.1 L
Remaining mass NaCl = 59.5 grams
Molar mass NaCl= 58.44 g/mol
Step 2: Calculate the dissolved mass of NaCl
100 - 59. 5 = 40.5 grams
Step 3: Calculate moles
Moles NaCl = 40.5 grams / 58.44 g/mol
Moles NaCl = 0.693 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity dissolved NaCl = 0.693 moles / 0.1 L
Molarity dissolved NaCl = 6.93 M
The molarity of the dissolved NaCl is 6.93 M
<u>Ionic Bond</u> is formed when the electronegativity difference is 0.4 > 2.0. Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
Explanation:
Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
An electronegativity of an atom is affected by
- The atomic number of the atom
- Secondly by the distance at which the valence electron are residing from the nucleus
1. In case the electronegativity difference (which is denoted by ΔEN) is less than 0.5 then the bond formed is known as N<u>onpolar covalent.
</u>
2. In case the ΔEN is in between 0.5 and 1.6, the bond formed is referred to as the<u> Polar covalent
</u>
3. In case the ΔEN is more /greater than 2.0, then the bond formed is referred to as<u> Ionic Bond</u>
<u>2 Examples of Ionic bonds</u>
- The formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom is an example of Ionic bond formation.
- Another example is the formation of NaCl from sodium (Na),which is a metal, and chloride (Cl), which is a nonmetal
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
