Answer:
The calculated volume of a lithium atom is 1.47 X 10⁻²⁹ m³.
Explanation:
Assuming the atom's volume is spherical (though it’s actually more complex), the volume is mathematically represented by:
----------------------------------------------------------------------------------------(Eq. 1)
Here, R corresponds to the lithium atom's radius. Since the radius is provided in picometers, we first convert it into meters.
Substituting this value into Eq.1 provides the desired result.
V= 1.47 X 10⁻²⁹ m³
Answer:
Explanation:
A mole is defined as the number of molecules divided by 6.02×10^23
Mole = (2×10^19)/(6.02×10^23)
Mole = 3.32×10^-5 mole
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.
Answer:
The original halide's formula is SrCl₂.
Explanation:
- The chemistry reaction's balanced equation is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X indicates the halide.
- Based on the equation's stoichiometry, 1.0 mole of strontium halide yields 1.0 mole of SrSO₄.
- The moles of SrSO₄ (n = mass/molar mass) = (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The moles of SrX can thus be calculated as 4.11 x 10⁻³ moles based on stoichiometry from the balanced equation.
- n = mass / molar mass, thus n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ is calculated using mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- Calculating the atomic mass of halide X, we find = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This identifies the atomic mass of Cl.
- Consequently, the original halide's formula is SrCl₂.
First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol:
3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10
Next convert moles to molecules using Avogadro’s number:
0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10
Each butane molecule contains 4 carbon atoms, so:
3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.
