Answer:
If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.
Explanation:
The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.
The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:
Vmix = 0.300*57 + 0.200*17
Vmix = 17.1 + 3.4
Vmix = 20.5 cm³
Answer:
The answer to your question is 80.3%
Explanation:
Data
Percent by mass of F
molecules NF₃
Process
1.- Calculate the molar mass of nitrogen trifluoride
molar mass = (1 x 14) + (19 x 3)
= 14 + 57
= 71 g
2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.
71 g of NF₃ ------------------ 100%
57 g of F ------------------- x
x = (57 x 100)/71
x = 5700 / 71
x = 80.3%
3.- Conclusion
Fluorine is 80.3% by mass of the molecule NF₃
Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
