1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
astraxan [27]
3 years ago
6

A student was performing a separation of a mixture of organic compounds. The final step of the process involved a filtration of

the analyte from an aqueous solution. After drying the filtered solid for a very short period time, they took the melting point of the compound. The measured melting point range of the compound was 106 – 113.8 0C, while the literature melting point of the compound was 122.3 0C. The above scenario is a very common one in organic labs.
1. Do you think their sample was pure?
2. If not, then what do you think could be the source of error.
3. How do you think this error can be minimized?
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

1) No

2) The solvent contaminated the analyte

3) The solvent should be evaporated properly before washing and drying the analyte

Explanation:

During separation of organic compounds, solvents are used. These solvents are able to contaminate the analyte and lead to a large difference in melting point of solids obtained.

However, the error can be minimized by evaporating the solvent before washing, drying and melting point determination of the solid.

You might be interested in
How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit
NARA [144]
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
5 0
4 years ago
Read 2 more answers
Guys someone shared their notes with me how should I thank them
Dvinal [7]
مسويها اصريت تصريحات
6 0
2 years ago
Read 2 more answers
I NEED HELP ASAP! PLEASE BE GENUINE
love history [14]

1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

M₁ = 0.096 g/mol

<h3>2. How to determine the pressure of O₂</h3>

From the question given above, the following data were obtained:

  • Volume (V) = 438 L
  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

P = 1.524 atm

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

6 0
3 years ago
6. A 25.0-mL sample of potassium chloride solution was found to have a mass of 25.225 g.
Morgarella [4.7K]

Answer:

6. a. 5.53 (m/m) %; b. 0.7490M

7. 0.156M

Explanation:

6. In the solution of KCl, there are 1.396g of KCl in 25.225g of solution. As mass/mass percent is:

Mass solute / Mass solution * 100

The mass/mass percent of KCl is:

a. 1.396g KCl / 25.225g solution * 100

5.53 (m/m) %

b. Molarity is moles of solute per liters of solution:

<em>Moles KCl -Molar mass: 74.55g/mol-:</em>

1.396g KCl * (1mol / 74.55g) = 0.018726 moles KCl

<em>Volume in liters: 25mL = 0.025L</em>

Molarity:

0.018726 moles KCl / 0.025L = 0.7490M

7. 0.90% means 0.90g of NaCl in 100g of solution:

<em>Moles NaCl -Molar mass: 58.44g/mol-:</em>

0.90g NaCl * (1mol / 58.44g) = 0.0154 moles NaCl

<em>Volume in Liters:</em>

100g * (1mL / 1.01g) = 99mL = 0.099L

Molarity:

0.0154 moles NaCl / 0.099L =

0.156M

4 0
3 years ago
How many atoms of sulfur, S, are in 4.00 g sulfur?
Angelina_Jolie [31]
<h3>Answer:</h3>

7.51 × 10²² atoms S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.00 g S

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.00 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S})
  2. Multiply:                            \displaystyle 7.51107 \cdot 10^{22} \ atoms \ S

<u>Step 4: Check</u>

<em>Follow sig figs and round. We are given 3 sig figs.</em>

7.51107 × 10²² atoms S ≈ 7.51 × 10²² atoms S

5 0
3 years ago
Other questions:
  • What is the number of carbon atoms in the ring portion of the haworth structure of glucose?
    15·1 answer
  • Electrons is an excited state are more likely to enter into chemical reactions.
    12·1 answer
  • Substance in mixtures do not change chemically. True or false
    10·1 answer
  • What pressure is required to reduce 57 mL of
    15·1 answer
  • A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
    10·1 answer
  • Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.
    13·1 answer
  • Balance each of the following equations by placing coefficients in front of the formulas as needed MgO(s) Mg(s) + 02(8) ZnCI.(ag
    15·1 answer
  • If the pressure is 0.9 kPa. and the volume is 4.0L, then the pressure changes to 0.20 kPa. What is the new volume?
    9·1 answer
  • Which enzyme of the citric acid cycle most closely resembles the pyruvate dehydrogenase complex in terms of its structure, organ
    15·1 answer
  • If you were to look up at the moon and see the shape of the moon as a circle, what would the next observable shape in the cycle
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!