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blsea [12.9K]
4 years ago
12

A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the ca

binet is used to store bottles of tetrahydrofuran, carbon tetrachloride, pentane, dimethyl sulfoxide, and acetone. The chemist plans to try to identify the unknown liquid by measuring the density and comparing to known densities. First, from his collection of Material Safety Data Sheets (MSDS), the chemist finds the following information: liquid density of tetrahydrofuran 0.89·gcm^−3, carbon tetrachloride 1.6·gcm^−3, pentane 0.63·gcm^−3, dimethyl sulfoxide 1.1·gcm^−3, acetone 0.79·gcm^−3. Next, the chemist measures the volume of the unknown liquid as 0.852L and the mass of the unknown liquid as 938.g . Calculate the density of the liquid. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
lord [1]4 years ago
8 0

Answer:

DMSO

Explanation:

Volume of unknown liquid = 0.852L = 852mL or 852cm^3

Mass of unknown liquid = 938g

Density of unknown liquid = Mass of liquid / Volume of liquid

                                            = 938 / 852 = 1.100g/cm^3

From the given other data, the unknown liquid is Dimethyl sulfoxide (DMSO) as the MSDS of the mentioned liquid has been given 1.1g/cm^3

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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