I think it is 110 mL but I’m not totally sure.
Answer:
1.3 L.
Explanation:
- Molarity is the no. of moles of solute per 1.0 L of the solution.
<em>M = (no. of moles of CaSO₄)/(Volume of the solution (L))</em>
<em></em>
M = 0.352 M.
no. of moles of CaSO₄ = mass/molar mass = (62.1 g / 136.14 g/mol) = 0.456 mol,
Volume of the solution = ??? L.
∴ (0.352 M) = (0.456 mol)/(Volume of the solution (L))
<em>∴ (Volume of the solution (L) </em>= (0.456 mol)/(0.352 M) = <em>1.296 L ≅ 1.3 L.</em>
Answer:
2 mole of C4H10 were used in this reaction.
Answer:
The empirical formula is P2O5 (option B)
Explanation:
An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.
The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.
The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.
If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5
To find the empirical formula we divide the molecular formula by 2 what will give us P2O5
For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5
This is the simpliest form for the compound P4O10.
The empirical formula is P2O5 (option B)
<span>13.7 g
The balanced formula is:
Pb(ClO3)2(aq)+2NaI(aq) ==> PbI2(s) + 2NaClO3(aq)
The number of moles of NaI we have is the volume of fluid multiplied by the molarity so
0.350 * 0.170 = 0.0595 moles
Since the NaI is the limiting reactant, for every two moles used, we'll produce 1 mole of precipitate. So
0.0595 mole / 2 = 0.02975 mole
Now we need to calculate the molar mass of PbI2. Looking up the atomic weights
Atomic weight Lead = 207.2
Atomic weight iodine = 126.90447
Molar mass PbI2 = 207.2 + 2 * 126.90447 = 461.00894 g/mol
Now multiply the molar mass by the number of moles we have.
461.00894 g/mol * 0.02975 mol = 13.71501597 g
Rounded to 3 significant figures, the answer is 13.7 g</span>