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pashok25 [27]
3 years ago
15

If I have 4.4 moles of nitrous oxide (laughing gas) that is kept at 32.9 °C in a container under 1.6 atm, what is the volume of

the container? (R = 0.0821 (L*atm)/(mol*K))
Round your answer to 1 decimal place.
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

Explanation:

Using the ideal gas equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information provided in this question;

P = 1.6atm

n = 4.4 moles

R = 0.0821 L*atm/mol*K

T = 32.9°C = 32.9 + 273 = 305.9K

V = ?

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0.000000655 mL in scientific notation
alina1380 [7]

Answer:

<u>The answer is 6.55 * 10⁻⁷</u>

Explanation:

Let's write 0.000000655 mL in scientific notation:

The first step is you have to move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the right in 0.000000655 you will get 6.55.

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<u>The answer is 6.55 * 10⁻⁷</u>

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4 years ago
Complete the word equation for making a salt.<br> Metal oxide +________<br> + salt + water
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Why does the chloride ion have a charge of -1
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8 0
3 years ago
8. In the lab, you are asked to prepare 150 mL of an 85 g/L solution and then use the original
finlep [7]

For the original solution,

Concentration = 85g/L = 85g/1000ml

Required Volume = 150ml

(a)

Concentration = Mass/Volume

⇒ 85g/1000ml = Mass of the solute/150ml

⇒ Mass of the solute = 85 x 150/1000

⇒ Mass of the solute = 12.75 grams

Hence, 12.75 grams of the solute is required to prepare the original solution.

(b)

Now,

Concentration of the original solution = 85g/L = M1

Required concentration = 50g/L = M2

Final Volume = 500 ml = V2

Required Volume from the original solution = V1 =?

As we know,

M1V1 = M2V2

Hence,

85g/L x V1 = 500 ml x 50g/L

⇒ V1 = 500 X 50 / 85

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Therefore, 294.11 ml of the original solution is required to make the diluted solution.

<h3>What is a Stock solution?</h3>
  • A stock solution is a very potent mixture. These solutions are quite helpful since we may take a small amount of the stock solution and dilute it to the required concentration.
  • These ready-to-use chemical reagents can be prepared quickly thanks to these stock solutions. It also aids in material conservation.
  • As a result, just the amount of stock and solvent required for the dilution process is consumed when utilizing a stock solution to low-concentrated solution.
  • Because we simply need to dilute the stock solution instead of preparing the solution using many reagents and complex procedures, it is also crucial to conserve storage space.
  • Additionally, it enhances experiment accuracy.
  • A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration.
  • Examples of typical stock solutions in laboratories are sodium hydroxide and hydrochloric acid. These play a critical role in creating titration-related solution preparations.

To learn more about the Stock solution, refer to:

brainly.com/question/28083950

#SPJ13

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2 years ago
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