Question

volume of the box in for full credit 11) CALCULATIONS AND PROBLEMS Show all ORSERIE SIGNIFICANT FIGURES RULES! in the correct number of What is the temperature recorded by the following thermometer significant figures? (5 ps) 1 inch 2. An unknown substance is soluble in alcohol and cyclohexane but not in water. It appears to melt between 50 and 55 degrees Celsius. When added to 500 ml. of water, a 2.30 sportion causes the volume to rise to 520 mL. What is the desity of this a nce? (10 pts)

Answer 1

1. 20 ℃

2. Volume of water = 50.0 mL

Volume of water + substance = Increased volume = 52.0 mL

=> Volume of substance = 52.0 - volume of water

=> Volume of substance = 52.0 - 50.0 = 2.0 mL

Now, mass of the substance taken = 2.30 g

Since, density = mass/volume

Density = 2.30 ÷ 2.0 g/mL = 1.15 g/mL

Now, since 2.30 has three significant figures and 2.0 has two significant figures (trailing zeros are significant when specified by precision), the final answer should have two significant figures.

Density = 1.2 g/mL

3. Mass of crucible + lid = 22.120 + 30.145 g = 52.265 g

Mass of crucible, lid and iron = 53.215 g

=> Mass of iron = mass of crucible, lid and iron - mass of crucible + lid

=> Mass of iron = 53.215 - 52. 265 g = 0.95 g

Also, # of moles of iron = 0.95g ÷ 56 g/mol = 0.017 mol

Now, mass of crucible, lid and iron sulfide = 54.032 g

=> Mass of iron sulfide = mass of crucible, lid and iron sulfide - mass of crucible, lid

=> Mass of iron sulfide = 54.032 - 52.265 g = 1.767 g

Mass of sulfur = mass of iron sulfide - mass of iron = 1.767 - 0.95 g = 0.817 g

=> # of moles of sulphur = 0.817g ÷ 32 g/mol = 0.026 mol

a.

Calculation for empirical formula

Element	mass	molar mass	# moles	Ratio of moles	Simplest ratio
Fe	0.95	56	0.017	0.017/0.017 = 1	2
S	0.817	32	0.026	0.026/0.017 = 1.5	3

So, the empirical formula is Fe₂S₃.

b. 2 Fe + 3 S $\rightarrow$ Fe₂S₃

4. Cu + 2 HNO₃ (excess) $\rightarrow$ Cu(NO₃)₂ + H₂

According to this balanced chemical equation, 1 mol of Cu gives 1 mol of copper nitrate

=> 63.5 g of Cu gives 187.56 g of copper nitrate

=> 1 g of Cu gives 187.56 ÷ 63.5 g of copper nitrate

=> 1 g of Cu gives 2.9537 g of copper nitrate

Since, mass of Cu taken = 2.09 g

=> 2.09 g of Cu gives 2.95 × 2.09 g of copper nitrate

Theoretical yield = mass of copper nitrate obtained = 6.17 g

Experimental (actual) yield = 3.40 g

Percentage yield = (actual yield ÷ theoretical yield) × 100

% yield = (3.40 ÷ 6.17) × 100 = 55.1 %