Explanation:
The rate at which gases effuse is inversely related to the square root of their molar masses.
In this case, half of the helium (1.5 L) passed through the membrane in 24 hours. Therefore, we can calculate the effusion rate of He gas as follows.
= 0.0625 L/hr
Given that the molar mass of He is 4 g/mol and for 
it is 32 g/mol.
Now,
= 2.83
Thus, the effusion rate of 
= 
Rate of = 0.022 L/hr.
This implies that 0.022 L of gas will effuse in one hour.
Consequently, to find the duration needed for 1.5 L of gas to effuse, we calculate as follows.
= 68.18 hours
Thus, we can conclude that it will require 68.18 hours for half of the oxygen to effuse through the membrane.
Response:
How can you differentiate a physical change from a chemical change?
Clarification:
Answer:
Refer to the explanation.
Explanation:
Formation reactions involve the creation of one mole of a compound from its elements in their standard states.
NaBr (s)
The equation for the standard formation is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
As per appendix C, the standard heat of formation for NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The equation for the standard formation is
S (s) + (3/2) O₂ (g) → SO₃ (g)
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ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The equation for the standard formation is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
I hope this is helpful!
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Answer:
The correct choice for your inquiry is option A, Argon.
Explanation:
Isotope Atomic mass Percent (%)
1 35.9675 0.337
2 37.9627 0.063
3 39.9624 99.6
To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)
Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)
Average atomic mass = 0.1212 + 0.0239 + 39.8025
Average atomic mass = 39.9476
Theoretical Atomic mass
a) Ar 39.95
b) K 39.10
c) Cl 35.45
d) Ca 40.08
Answer:
The forward reaction will keep occurring until all NO or all NO₂ is consumed.
Clarification:
- According to Le Châtelier's principle, when a system at equilibrium experiences a disturbance from an outside source, the system will adjust to counteract this disturbance and restore equilibrium.
- Thus, removing the product (N₂O₃) from the system effectively lowers the product concentration, prompting the reaction to shift forward and generate additional product in order to alleviate the strain caused by the removal of N₂O₃.
- Consequently, the reaction will proceed forward until all of either NO or NO₂ is depleted.
