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brilliants [131]

3 years ago

13

A weather balloon is released in Houston when the pressure is 758.0 mmHg and 32.7°C. After the balloon reaches an altitude of 60

00 m it measures a volume of 425 L with a pressure of 364.2 mmHg and a temperature is 3.10°C. What was the initial volume of the balloon? Show all your work and round your answer to the hundredths place.
A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. The total pressure of the mixture is 100.0 kPa. If the partial pressure for helium is 83.1 kPa and 0.1000 kPa for carbon dioxide, what is the partial pressure of oxygen? Show all your work and round your answer to the tenths place.

1 answer:

erica [24]3 years ago

4 0

Answer:

See explanation

Explanation:

From the information provided;

P1 = initial pressure = 758.0 mmHg

T1 = Initial temperature = 32.7°C + 273 = 305.7 K

V1 = Initial volume?

P2 = Final pressure = 364.2 mmHg

V2 = Final volume = 425 L

T2= Final Temperature = 3.10°C + 273 = 276.1 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V1 = P2V2T1/P1T2

V1 = 364.2 * 425 * 305.7/758.0 * 276.1

V1 = 226. 18 L

B) Total pressure = 100.0 kPa

Pressure of Helium = 83.1 kPa

Pressure of carbon dioxide = 0.1000 kPa

Pressure of oxygen ?

From Dalton's law of partial pressures;

PT = P1 + P2 +P3 +.........

Hence;

Poxygen = 100.0 kPa - (83.1 kPa + 0.1000 kPa)

Poxygen = 16.8 kPa

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

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Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
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Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

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The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

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<h3>Answer:</h3>

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<u>Math</u>

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Parenthesis
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Addition
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