What pressure in Pascals will be exerted by 4.78 grams of oxygen gas in a 2.5-liter container at 20 C?

Answer 1

#P~=150*kPa#

From the Ideal Gas equation..........

#P=(nRT)/V=((4.78*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx293*K)/(2.50*L)#
#=1.44*atm-=1.44*atmxx101.3*kPa*atm^-1=145.6*kPa#
Note that you SIMPLY MUST KNOW that #H_2, O_2, N_2, F_2, Cl_2# are binuclear as the elements. In fact all the elemental gases (at RT) EXCEPT for the Noble Gases are bimolecular.
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Answer 2

The pressure exerted by 4.78 grams of oxygen gas in a 2.5-liter container at 20°C can be calculated using the ideal gas law equation (PV = nRT), where (P) is pressure, (V) is volume, (n) is the number of moles of gas, (R) is the ideal gas constant, and (T) is temperature in Kelvin.

First, calculate the number of moles of oxygen gas using the formula (n = \frac{m}{M}), where (m) is the mass of the gas and (M) is the molar mass of oxygen gas (O2). Then, convert the temperature from Celsius to Kelvin by adding 273.15.

After finding the number of moles, plug in the values into the ideal gas law equation to solve for pressure ((P)).

(n = \frac{m}{M}) (n = \frac{4.78 , \text{g}}{32 , \text{g/mol}}) (n ≈ 0.149 , \text{mol})

(T = 20 , \text{°C} + 273.15 = 293.15 , \text{K})

(PV = nRT) (P \times 2.5 , \text{L} = 0.149 , \text{mol} \times 8.31 , \text{J/(mol}\cdot\text{K)} \times 293.15 , \text{K}) (P \times 2.5 , \text{L} = 2912.3 , \text{J}) (P = \frac{2912.3 , \text{J}}{2.5 , \text{L}}) (P ≈ 1164.92 , \text{Pa})

So, the pressure exerted by 4.78 grams of oxygen gas in a 2.5-liter container at 20°C is approximately 1164.92 Pascal (Pa).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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