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Xelga [282]
3 years ago
12

He partial pressure of oxygen gas in our atmosphere is 0.21 atm. this is the partial pressure at which human lungs have evolved

to be able to breathe this gas. a scuba diver, will thus still have to breath oxygen at this pressure even when diving way down in the water. if a mixture of helium and oxygen (heliox) in his tank is at a pressure of 8.30 atm, what must the partial pressure be of helium to keep the partial pressure of oxygen at 0.21 atm?
Chemistry
1 answer:
solong [7]3 years ago
6 0
Hello!

The partial pressure of helium to keep the partial pressure of oxygen at 0,21 atm in a scuba-diver tank is 8,09 atm

To solve this question, we can use the Dalton's Law, which states that the total pressure in a container with a mixture of gases is the sum of the partial pressures o each individual gas. For the case of this mixture the Dalton's Law is as follows:

P_{tot}=P_{He} +P_{O_2}

In this equation, we need to clear for PHe, knowing that the PO₂ should be 0,21 atm, to find the required pressure of Helium:

P_{He}=P_{tot} -P_{O_2}=8,30atm-0,21atm=  8,09 atm

Have a nice day!
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A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.
Marina CMI [18]
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is  the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.
7 0
3 years ago
At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz
Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P°    (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles:   n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

<h2>P(C₆H₆) = 0.2961 atm</h2>

Hope this helps

7 0
2 years ago
Bromine has two isotopes, Br79 and Br81. The isotopes occur in a 50:50 (1:1) ratio. Given that the mass spectrum of bromine cont
AlekseyPX

Answer:

There will be 3 peaks.

Relative height of the atomic peaks would be; 158, 160 and 162

Explanation:

We are told that Bromine has two isotopes namely 79Br and 81Br in a 1 : 1 ratio (50 : 50).

This means that a compound which contains 1 bromine atom will have two peaks in the molecular ion region but it depends on which bromine isotope is contained in the molecular ion.

Thus;

Relative height of atomic peaks is given by;

m/z = 79Br¯ 79Br+ = 158

79Br¯ 81Br+ = 160

81Br¯ 81Br+ = 162

7 0
3 years ago
Can someone help me please .
lyudmila [28]

Sure. Whith what do you need help?

4 0
3 years ago
How is alkyne converted to acetone?
cricket20 [7]

Answer:

Propyne can be converted to the acetone when it is made to undergo the reaction with mercuric sulphate followed by hydrolysis and thus, resultant product, thus formed is acetone.

Explanation:

3 0
3 years ago
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