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1 month ago

Convert 338 L at 63.0 atm to its new volume at standard pressure.

Chemistry

1 answer:

lions [2.9K]1 month ago

7 0

The new volume at a standard pressure of 1 atm is 21294 liters.

Explanation:

Provided data:

Initial volume of the gas V1 = 338 liters

Initial pressure on the gas P1 = 63 atm

Standard pressure P2 = 1 atm

Final volume at standard pressure V2 =?

The equation of Boyle's law is applied here:

P1V1 = P2V2

To find V2, rearranging gives us:

V2 = $\frac{P1V1}{P2}$

Substituting the values:

V2 = $\frac{338X63}{1}$

= 21294 L

When the pressure is lowered to 1 atm, the gas volume expands significantly to 21294 liters.

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You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [2795]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

7 0

1 month ago

A gas sample of argon, maintained at constant temperature, occupies a volume of 500. l at 4.00 atm. what is the new volume if th

lorasvet [2795]

Boyle's law describes the relationship between gas pressure and volume.
It asserts that at a constant temperature, pressure is inversely proportional to gas volume.
PV = k
where P represents pressure, V denotes volume, and k is a constant.
P1V1 = P2V2
where the parameters for the initial condition are on the left, and the parameters for the second condition appear on the right side of the formula.
By substituting values into the equation: 4.00 atm x 500 L = 8.0 atm x V
V calculates to 250 L.
Thus, the new volume becomes 250 L.

6 0

1 month ago

Read 2 more answers

A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of

Tems11 [2777]

1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.

This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.

2) Molarity is represented as M = moles of solute / liters of solution.

Thus, we need to determine the moles in 8.9 grams of NaCl.

3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.

4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.

5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.

Answer: 0.152 M

4 0

1 month ago

Acetone major species present when dissolved in water

Alekssandra [3086]

The compound is acetone ( CH₃-CO-CH₃)

Explanation:

1) Acetone is represented as CH₃-CO-CH₃.

2) This is a molecule formed by covalent bonds.

3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.

In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.

This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).

3 0

1 month ago

KiRa [2933]

Explanation:

It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.

$\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$

Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.

4 0

1 month ago