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
IRISSAK [1]
2 years ago
14

A student dissolves 7.9 g of stilbene (C14H12) in 475. mL of a solvent with a density of 1.03 g/mL. The student notices that the

volume of the solvent does not change when the stilbene dissolves in it. Calculate the molarity and morality of the students solution. Round both of your answers to 2 significant digits.
Chemistry
1 answer:
andrey2020 [161]2 years ago
6 0

Answer:

Molarity: 0.092M

Molality: 0.090m

Explanation:

Molarity, M, is defined as the moles of solute (In this case, C14H12 -Molar mass: 180.25g/mol-) in 1L of solution.

The molality, m, are moles of solute per kg of solvent.

<em>Molarity:</em>

<em>Moles solute: </em>

7.9g * (1mol/180.25g) = 0.04383 moles

<em>Liters solution:</em>

475mL = 0.475L

Molarity: 0.04383 moles / 0.475L = 0.092M

<em />

<em>Molality:</em>

<em>kg solvent:</em>

475mL * (1.03g/mL) = 489.25g = 0.48925kg

Molality:

0.04383 moles / 0.48925kg = 0.090m

You might be interested in
When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of
jasenka [17]
2Al + 3ZnCl₂ ----> 3Zn + 2AlCl₃

What is the coefficient in front of AlCl₃? ==>> 2
4 0
3 years ago
1. Which of the following statements is a consequence of the equation E=MC2
finlep [7]

Answer:

D. All of the above​

Explanation:

E = MC² is a common equation in physics.

E is energy

M is mass

C is the speed of light

The law was stated by Albert Einstein.

  • From this law, it was shown that energy is released when matter is destroyed.
  • Mass and energy are equivalent as seen in nuclear reactions where mass is converted to energy.
  • Mass and energy is usually conserved in any process and this is a subtle modification of the law of conservation of matter and energy.
  • Most of these postulates apply to nuclear reactions which generally do not follow some precepts of chemical laws.
7 0
3 years ago
Se valoró una alícuota de 10ml de HCl con NaOH AL 0.0548M de la cual se gastó 17.9ml. La concentración molar del ácido es:
Ymorist [56]

Answer:

se lonh hion ffyuk jlngvn

7 0
2 years ago
Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =
Viktor [21]

Answer:

The answer is 5.10

Explanation:

<h3><u>Given</u>;</h3>
  • pH = 1.9
<h3><u>To </u><u>Find</u>;</h3>
  • pOH = ?

We know that

pH + pOH = 7

pOH = 7 – pH

pOH = 7 – 1.90

pOH = 5.10

Thus, The pOH of the solution is 5.10

6 0
1 year ago
Use the following graph of a car traveling on a straight northerly path to answer this question. At what time would the
BARSIC [14]

Answer:

B 144.0 s is the best answer of this question

6 0
3 years ago
Other questions:
  • A solution has a concetration of 0.3mol/dm3 of sodium hydroxide .what volume of
    5·1 answer
  • Help plz this was due 2 days ago but I don’t understand it at all
    5·2 answers
  • How does the structure of covalent bond affect the structure of covalent compounds
    6·1 answer
  • A/an _______ bond allows metals to conduct electricity
    6·2 answers
  • If you have a sample of honey that has a mass of 20 grams what is the volume
    9·1 answer
  • The electron configuration of a neutral atom is 1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons.
    6·1 answer
  • The diagram is a model of one way that materials move into a cell.
    7·2 answers
  • If you had an aqueous mixture that contained Ag+ , K+ , and Pb2+ cations, how many different solids could precipitate if a chlor
    9·1 answer
  • 4Fe + 302 → 2Fe2O3
    14·1 answer
  • Picture of the question is below, if answered correctly, I will reward with brainliest.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!