Answer:
Indeed, the chemist is capable of identifying the compound present in the sample.
Explanation:
In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.
For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.
When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.
- If the mass ratio is 0.830, then it indicates a pure K₂O compound.
- If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
- If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.
Answer: The correct selection is (b).
Explanation:
The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.
This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.
Therefore, a significant amount of energy is needed to dislodge the valence electrons.
The electronic configuration for helium is
. Hence, due to its fully occupied valence shell, it exhibits greater stability.
Consequently, a large amount of energy is needed to remove an electron from a helium atom.
In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.
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.
Answer:
The nichrome wire has contaminants.
The sample solution might be tainted.
Explanation:
If the nichrome wire is contaminated, sodium impurities could be causing the yellow flame. The wire is initially placed in the flame without the sample to check for such impurities.
The testing solution could also be contaminated, causing it to display a color different from the anticipated shade of the test ion.
Answer :
The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.
Percent Ionic Character:
This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Here, Xa is the electronegativity of atom A and Xb is that of atom B.
Given: The compound is TiAl₃.
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 (as shown in the provided image)
Substitute these values into the formula:



The value of e⁻¹ equals 0.90.
Therefore, percent ionic character = (1 - 0.90) × 100
Percent Ionic Character = 10%
Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.