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blsea [12.9K]
3 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]3 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

You might be interested in
What are the empirical formula and empirical formula mass for C10H30O10?
AysviL [449]

Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

As

C₁₀H₃₀O₁₀ Consist of  10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

                         C : H : O

                        10 : 30 : 10

Divide the ratio by two to get simplest ratio

                          C      :   H      :    O

                         10/10 : 30/10 : 10/10

                             1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

                     Empirical formula of C₁₀H₃₀O₁₀ =  CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

4 0
3 years ago
Calculate the volume of liquid in the tank sketched below. Give your answer in liters, and round to the nearest 0.1 L.
tensa zangetsu [6.8K]

Answer:

The answer is 18.9

Explanation:

3 0
3 years ago
A local am radio station broadcasts at a wavelength of 349m, calculate the energy of the wavelength at which it is broadcasting?
Juli2301 [7.4K]

Answer:

E = 5.69x10⁻²⁸m

Explanation:

To solve this question we neeed to convert the wavelength in meters to energy in joules using the equation:

E = hc / λ

<em>Where E is energy in joules, h is Planck's constant = 6.626x10⁻³⁴Js</em>

<em>c is light constant = 3.0x10⁸m/s</em>

<em>And λ is wavelength in meters = 349m</em>

Replacing:

E = 6.626x10⁻³⁴Js*3.0x10⁸m/s / 349m

E = 5.69x10⁻²⁸m

5 0
3 years ago
Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system,
givi [52]

Answer: The order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1 

B)-61.9 kJ

Explanation:

The change in the internal energy of a system  is positive if the reaction absorbs energy and  negative if the reaction releases energy. For a system to cause an increase in volume, it must have very high energy built up to be released.

1. Surroundings get colder and the system decreases in volume. Here, the surrounding absorbs energy  resulting in positive  ΔE

2. Surroundings get hotter and the system expands in volume.  Here energy is released causing the system to be negative

3. Surroundings get hotter and the system decreases in volume. Although there is a decreased volume, the system is negative because it releases energy

4. Surroundings get hotter and the system does not change in volume.  System is negative because it releases energy even thgoygh there is no change in volume

Therefore the order from the Most energy released to most Energy   Absorbed Is given as  2---> 4--->,3-->---> 1

b) Using  

 ΔE = q+ w     from 1st law of thermodynamics

 ΔE=  ΔH - P  ΔV

gIven  

 ΔH = -75.0KJ

volume=  A change  from 5.0L TO 2.0L = Final volume - initial volume = 2-5= -3.00L

P= 43.0atm

ΔE=  ΔH - P  ΔV

P  ΔV  = 43 atm x -3 = -129L.atm

We first convert  L-atm to Joules.

1 L-atm = 101.325 Joules.  

129L.atm = 129 x 101.325 = - 13071 J

to KJ becomes

13071/1000 = - 13.071KJ

Recall ΔE=  ΔH - P  ΔV and putting values

ΔE  = -75.0 - (-13.071 KJ)= -75.0 kJ + 13.071 kJ = -61.9 kJ

8 0
3 years ago
Which process is an oxidation?
Zarrin [17]
<span>The problem has to do with oxidation states of the matter. The oxidation state of oxygen will always be -2 with the exception of peroxides which will have a state of -1. The overall balanced state of chemical compounds will be 0, so the oxidation state of Mn in MnO2 will be +4. The oxidation state of MnO4- will then be +7 to balance out to the negative one charge. The state change from +4 to +7 is 3, thus three electrons have to be lost in order for this to happen; a loss of a charge of -3 results in an increase of charge of 3. Oxidation is always the process of 'losing' electrons.

</span><span>E] MnO2(s) MnO4-(aq</span>
4 0
3 years ago
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