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2 years ago

How many moles of propane (C3H8) are in a 7.00-L tank at 20.0°C and 5.45 atm of pressure?

1 answer:

7 0

By the claperyron equation<span>:

T = 20.0ºC + 273.15 => 293.15 K

R = 0.082 atm

Therefore:

P x V = n x R x T

5.45 x 7.00 = n x 0.082 x 293.15

38.15 = n x 24.0383

n = 38.15 / 24.0383

n = 1.587 moles of C3H8

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How many moles of gas would occupy 22.4 L at 273K in one arm?

andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law 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)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

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1 year ago

How do you solve this ??

kotegsom [21]

Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

$E_{eV} = E_{J} X (6.241 X 10^{18})$

Here, we have the value of particle in terms of Joules which is 3.2 X $10^{-19}$

So, on substituting we get,
$E_{eV}$ = 3.2 X $10^{-19}$ X $(6.241 X 10^{18} )$

$E_{eV}$ = 1.99 eV so, it can be rounded off to 2.00 eV.

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2 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

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2 months ago

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xxTIMURxx [149]

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