Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P× V = n× R× T

In this case, you know:

P= 2 atm
V= ?
n= $9.49 gramsx\frac{1 mole}{2 grams} = 4.745moles$ being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole.
R= 0.082 $\frac{atmL}{molK}$
T= 353 K

Replacing:

2 atm× V = 4.745 moles× 0.082 $\frac{atmL}{molK}$ × 353 K

Solving:

V = (4.745 moles× 0.082 $\frac{atmL}{molK}$ × 353 K)÷ 2 atm

Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.

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5 0

3 years ago

The mixing of several atomic orbitals to form the same number of equivalent hybrid orbitals

Temka [501]

Answer:

Hybridization

Explanation: -

Hybridization occurs when atomic orbitals mix to form a new atomic orbital

3 0

3 years ago

At normal blood pH pH (7.4), hemoglobin is 80 80 % saturated at a partial pressure of oxygen ( O 2 O2 ) of 40 mmHg 40 mmHg . Use

schepotkina [342]

Answer:

An example of oxygen–hemoglobin (O2–Hb) dissociation curves from (A) one penguin at pH 7.5, 7.4 and 7.3, and (B) the emperor penguin, the bar-headed goose (Anser indicus) (Black and Tenney, 1980) and the domestic duck (Anas platyrhynchos, forma domestica) (Hudson and Jones, 1986) at pH 7.4. Note that as for the bar-headed goose, the O2–Hb dissociation curve of the emperor penguin is significantly left-shifted as compared with the domestic duck (and most birds). The bar-headed goose photo is courtesy of Graham Scott; the domestic duck photo is by Maren Winter (licensed under the terms of the GNU Free Documentation License, Version 1.2 or any later version); the penguin photo is by J.M.

Explanation:

The resulting regression equations from the plots of log[SO2/(100–SO2)] vs log(PO2) (all saturation points, all penguins combined) were:

pH 7.5: log[SO2/(100–SO2)] = 2.92589 × log(PO2) – 4.24338 (N=43, r2=0.98, P<0.0001),

pH 7.4: log[SO2/(100–SO2)] = 2.94767 × log(PO2) – 4.39858 (N=70, r2=0.98, P<0.0001),

pH 7.3: log[SO2/(100–SO2)] = 3.04945 × log(PO2) – 4.72019 (N=38, r2=0.99, P<0.0001),

pH 7.2: log[SO2/(100–SO2)] = 3.15958 × log(PO2) – 4.97618 (N=9, r2=0.99, P<0.0001).

5 0

4 years ago