The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.
<h3>How do we calculate the partial pressure of gas?</h3>
Partial pressure of particular gas will be calculated as:
p = nP, where
- P = total pressure = 748 mmHg
- n is the mole fraction which can be calculated as:
- n = moles of gas / total moles of gas
Moles will be calculated as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole
Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole
Mole fraction of hydrogen = 1 / (1+0.5) = 0.6
Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm
Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.
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Stored energy is described as potential energy
Answer:
Helum (He)g will escape faster
Explanation:
the phenomemenon can be explained by the Graham's law of diffusion.
Graham's law of difussion states that the rate of difussion is inversely proportional to the square root of the molecular mass,which means the gas with lower molecular mass will escape faster.
Helium gas has a molecular mass of 4 while Neon has a molecular mass of 10.
rate of diffusion of He/rate of difussion of Ne=√4/10=√0.4=0.63
It means He(g) will move 0.63 times faster than Ne(g) under the same condition
Answer:
the solubility increases
Explanation:
The solubility of any ionic compound is due to the ionization of the compound and then the strong ion-dipole interactions acting between the ions and the solvent.
Thus, solubility also depends on the extent of the ionization of the salt.
The more the salt ionizes, the more there is ion-dipole interaction between the ions the solvent and more is the solubility.
Answer:
464.1 J absorbed.
Explanation:
Given data:
Specific heat of zinc = 0.39 J/g°C
Mass of zinc = 34 g
Temperature changes = 22°C to 57°C
Energy absorbed or released = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57°C - 22°C
ΔT = 35°C
Q = m.c. ΔT
Q = 34 g. 0.39 J/g°C. 35°C
Q = 464.1 J