Answer:
- m = 1,000/58.5
- b = - 1,000 / 58.5
1) Variables
- molarity: M
- density of the solution: d
- moles of NaCl: n₁
- mass of NaCl: m₁
- molar mass of NaCl: MM₁
- total volume in liters: Vt
- Volume of water in mililiters: V₂
- mass of water: m₂
2) Density of the solution: mass in grams / volume in mililiters
3) Mass of NaCl: m₁
Number of moles = mass in grams / molar mass
⇒ mass in grams = number of moles × molar mass
m₁ = n₁ × MM₁
4) Number of moles of NaCl: n₁
Molarity = number of moles / Volume of solution in liters
M = n₁ / Vt
⇒ n₁ = M × Vt
5) Substitue in the equation of m₁:
m₁ = M × Vt × MM₁
6) Substitute in the equation of density:
d = [M × Vt × MM₁ + m₂] / (1000Vt)
7) Simplify and solve for M
- d = M × Vt × MM₁ / (1000Vt) + m₂/ (1000Vt)
- d = M × MM₁ / (1000) + m₂/ (1000Vt)
Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water means 1000Vt = V₂
- d = M × MM₁ / (1000) + m₂/ V₂
m₂/ V₂ is the density of water: 1.00 g/mL
- d = M × MM₁ / (1000) + 1.00 g/mL
- M × MM₁ / (1000) = d - 1.00 g/mL
- M = [1,000/MM₁] d - 1,000/ MM₁
8) Substituting MM₁ = 58.5 g/mol
- M = [1,000/58.5] d - [1,000/ 58.5]
Comparing with the equation Molarity = m×density + b, you obtain:
- m = 1,000/58.5
- b = - 1,000/58.5
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Explanation:
A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.
For example, melting point of water is 
. So, at this temperature solid state of water and liquid state are present in equilibrium with each other.
Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is 
then some of the solid will change into liquid state but the temperature will remains the same.
Answer:
The answer to this would be communicating.
Explanation:
A scientist would be communicating to his or her fellow colleagues and sharing to them his or her idea.
Hope you find this answer helpful! :)
Aobt 1,200 halp you at all
