Answer:
%Li-6 = 5.996% & %Li-7 = 94.004%
Explanation:
let X₁ = Li-6 & X₂ = Li-7 where Xₙ = mole fraction
X₁ + X₂ = 1 => X₁ = 1 - X₂
6·X₁ + 7·X₂ = 6.94004
=> 6(1 - X₂) + 7·X₂ = 6.94004
=> 6 - 6·X₂ + 7·X₂ = 6.94004
=> 6 + X₂ = 6.94004
X₂ = 6.94004 - 6 = 0.94004 => %X₂ = %Li-7 = 94.004%
X₁ = 1.00000 - 0.94004 = 0.05996 => %X₁ = %Li-6 = 5.996%
Answer:
There are
Explanation:
There are 10 elements in CH3CH2COONa.
The electron group arrangement of CH₂Cl₂ is penta-atomic. The molecular shape is tetrahedral, and the bond angle is 109.5°.
<h3>What is the bond angle?</h3>
The bond angle is the angle between the atoms of the compound. The bond angle is defined in degree. The bond length is also there. It is the distance between the nuclei of the two atoms.
The bond angle of CH₂Cl₂ is 109.5°. According to the VSEPR theory, as the structure will be tetrahedral.
Thus, the penta-atomic configuration of CH2Cl2's electrons. Tetrahedral is the molecular shape, the bond angle is 109.5°.
To learn more about bond angle, refer to the link:
brainly.com/question/14089750
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Answer:
108.43 grams KNO₃
Explanation:
To solve this problem we use the formula:
Where
- ΔT is the temperature difference (14.5 K)
- Kf is the cryoscopic constant (1.86 K·m⁻¹)
- b is the molality of the solution (moles KNO₃ per kg of water)
- and<em> i</em> is the van't Hoff factor (2 for KNO₃)
We <u>solve for b</u>:
- 14.5 K = 1.86 K·m⁻¹ * b * 2
Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:
- 275 mL water ≅ 275 g water
- moles KNO₃ = molality * kg water = 3.90 * 0.275
- moles KNO₃ = 1.0725 moles KNO₃
Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:
- 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
To find the molarity of the compound, simply determine the molar mass of MgCl2 and then convert 50 g to moles using the molar mass of the compound. Then convert 150 ml to L = 0.15 L
Then divide the moles amount by the volume in L.
