Given the balanced equation:
( Reaction type : double replacement)
CaF2 + H2SO4 → CaSO4 + 2HFI
We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.
This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.
For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.
For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.
Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.
Molality=mol/kg
342/171
=2m
Hope this helped :)
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Answer:
0.171 M
Explanation:
Step 1: Given data
- Mass of H₃PO₄ (solute): 3.35 g
- Volume of solution (V): 200 mL
Step 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
Answer:
1.51367e+10 inches
Explanation:
1 mile = 63360
63360 x 238900 = 15136704000
Hope this helped!
