I think the charge is +2
Hope this helps
Answer:
An ionic compound is formed when there is a reaction between the elements whose ions are electrostatically attracted.
Explanation:
The ionic compounds form crystalline networks with ionic bonding. Electrostatic attraction is a very strong bond that is very difficult to break. The stability of the ionic compound depends on the lattice energy, the higher it is, the more stable the compound is and the lattice energy is that which is released in the formation. At room temperature they are always in a solid state, because the bonds are very close and as stated before, they are difficult to break (providing a lot of energy)
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
Answer:
radius = 156 pm
Explanation:
The relation between radius and edge length of unit cell of BCC is
r=a
/4
Given
a = 360 pm
Therefore
r = r = radius = 360
/4= 155.88 pm
Or
156 pm
<u>Answer:</u> The freezing point of solution is 5.35°C
<u>Explanation:</u>
The equation used to calculate depression in freezing point follows:

To calculate the depression in freezing point, we use the equation:

Or,

where,
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point elevation constant = 4.90°C/m
= Given mass of solute (naphthalene) = 2.60 g
= Molar mass of solute (naphthalene) = 128.2 g/mol
= Mass of solvent (benzene) = 675 g
Putting values in above equation, we get:

Hence, the freezing point of solution is 5.35°C