Answer:
Option B. 176g/mol
Explanation:
We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:
Hexasodium means 6 sodium atom
Difluoride means 2 fluorine atom.
Therefore, the formula for hexasodium difluoride is Na6F2.
The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.
Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:
Molar mass of Na = 23g/mol
Molar mass of F = 19g/mol
Relative formula mass Na6F2 = (23x6) + (19x2)
= 138 + 38
= 176g/mol
Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol
Answer:
- What is the limiting reactant?: HCl is the limitng reactant.
- How many moles of H₂ are formed?: 6.5 moles of H₂ are formed.
Explanation:
Part A: <em>what is the limiting reactant?</em>
1) <u>Balanced chemical equation</u>: given
- 2Al + 6HCl → 2AlCl₃ + 3H₂
2)<em> </em><u>Stoichiometric mole ratio:</u>
Use the coefficients of the balanced equation:
- 2 mol Al : 6 mol HCl : 2 mol AlCl₃ : 3H₂
3) <u>Compare the stoichiometric mole ratio of the reactants with their actual ratio</u>:
- Theoretical ratio: 2 mol Al / 6 mol HCl ≈ 0.33 mol Al / mol HCl
- Actual ratio: 6.0 mol Al / 13 mol Cl ≈ 0.46 mol Al / mol Cl
Since the actual ratio indicates that there is a greater number of moles of Al (0.46) per mol of Cl than what is required by the stoichiometric ratio(0.33), Al is in excess and HCl is the limiting reactant.
Answer: the limiting reactant is HCl.
Part B. <em>How many moles of H₂ are formed?</em>
3. <u>Determine how many moles of H₂ can be formed</u>
- Theoretical ratio using limiting reactant:
6 mol HCl / 3 mol H₂ = 13 mol HCl / x
⇒ x = 13 mol HCl × 3 mol H₂ / 6 mol HCl = 6.5 mol H₂.
The answer must be reported with two significant digits, such as the data are given.
Answer: 6.5 moles of H₂ are formed
Concentration (m) signify 'molality'
Concentration (M) signify 'Molarity'
Formula for molality (m) =
In the question denisty of solute is not given so we can calculate concentration (M) that is 'Molarity'
Molarity (M) =
Moles of solute CH3OH = Given grams / Molar mass of CH3OH
Given grams of CH3OH = 12.9 g and molar mass = 32.0 g/mol
Moles of solute CH3OH =
Moles of solute CH3OH = 0.403 mol
Volume of solution = 230 mL , we need to convert 230 mL to 'L'
Volume =
Volume = 0.23 L
Molarity (M) =
Concentration (M) = 1.75 mol / L or 1.75 M
Answer:
Transition metals
Explanation:
Columns one and two are part of the main group along with columns 13 through 18