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Zanzabum
3 years ago
6

What is the vapor pressure of carbon disulfide at its normal boiling point?

Chemistry
1 answer:
mixer [17]3 years ago
7 0
I am not so sure, but based on what I'm seeing on the periodic table, the answer should be 3,827 degrees Celsius.
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Energy transformations are constantly taking place. <br><br><br> True or False
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The statement is : True 
 
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3 years ago
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The gas in an aerosol can is at a pressure of 3.16 atm at 32.2°C. What would the gas pressure in the can be at 22.9°C?
olasank [31]

Answer:

The pressure of the gas would be 3.06 atm

Explanation:

Amonton's law states that the pressure is directly proportional to the absolute temperature of a gas under constant volume. The equation is:

P1 / T1 = P2 / T2

<em>Where P1 is the initial pressure = 3.16atm</em>

<em>T1 is initial absolute temperature = 273.15 + 32.2°C = 305.35K</em>

<em>P2 is our incognite</em>

<em>And T2 is = 273.15 + 22.9°C = 296.05K</em>

<em />

Replacing:

3.16atm / 305.35K = P2 / 296.05K

3.06 atm = P2

<h3>The pressure of the gas would be 3.06 atm</h3>
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3 years ago
What is the name of this compound?<br><br> C6H6-<br> C=O-<br> H
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What is the density of a 12-gram sample of zinc with volume of 8.4 cm3? <br><br> pls hurry pls
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3 years ago
You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL
pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

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3 0
3 years ago
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