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Sedbober [7]
3 years ago
7

10mL volume of 75mM EtOH is necessary for an experiment. What volumes of 1M EtOH and water should be mixed to obtain the working

concentration?
Chemistry
1 answer:
kenny6666 [7]3 years ago
3 0

Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.

Explanation:

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1M

V_1 = volume of stock solution = ?

C_2 = molarity of diluted solution = 0.075 M    (1mM=0.001M)

V_2 = volume of diluted solution = 10 ml

Putting in the values we get:

1M\times V_1=0.075M\times 10ml

V_1=0.75ml

Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.

Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration

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[15 Points, Stoichiometry and Gases]
KiRa [710]

The reaction is

CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g) ​

As we have data of gas ethyne (or acetylene), C₂H₂

We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken

the moles of acetylene will be calculated using ideal gas equation

PV =nRT

R = gas constant = 0.0821 Latm/molK

T = 385 K

V = volume = 550 L

P = Pressure = 1.25 atm

n = moles = ?

n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol

As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide

moles of calcium carbide = 21.75mol

molar mass of CaC₂ = 40 + 24 = 64

mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g

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4 years ago
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Describe one trait that you believe is unique to you that you think is
gladu [14]

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Mom and dad have blue eyes but I have brown.

Explanation:

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3 years ago
The volume of a gas is 550 mL at 960 mm Hg and 200.0 C. What volume
Masja [62]

Answer:

The volume will be 568.89 mL.

Explanation:

Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Where P = pressure, T = temperature, K = Constant

Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Studying an initial state 1 and a final state 2, it is fulfilled:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 960 mmHg
  • V1= 550 mL
  • T1= 200 C= 473 K (being 0 C=273 K)
  • P2= 830 mmHg
  • V2= ?
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Replacing:

\frac{960 mmHg*550 mL}{473K} =\frac{830 mmHg*V2}{423 K}

Solving:

V2=\frac{423 K}{830 mmHg} *\frac{960 mmHg*550 mL}{473K}

V2= 568.9 mL

<u><em>The volume will be 568.89 mL.</em></u>

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3 years ago
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Answer:

V_2=0.046L

Explanation:

Hello,

In this case, we apply the Boyle's law as an inversely proportional relationship allowing us to understand the pressure-volume behavior as shown below:

P_1V_1=P_2V_2

In such a way, solving for the final volume V2, we obtain:

V_2=\frac{P_1V_1}{P_2}=\frac{0.2L*176mmHg}{760mmHg}  \\\\V_2=0.046L

Best regards.

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