Okay, I can explain the concepts behind those multiple-choice questions in more depth:
14. Ionic Compounds: Ionic compounds are formed through the transfer of electrons between atoms, typically between a metal and a nonmetal. This transfer creates ions (charged particles) that are attracted to each other due to their opposite charges.
- A) SF6 (Sulfur hexafluoride): This is a covalent compound because it consists of two nonmetals (sulfur and fluorine) sharing electrons.
- B) H2O (Water): This is also a covalent compound. Hydrogen and oxygen share electrons.
- C) H2O2 (Hydrogen peroxide): Another covalent compound where hydrogen and oxygen share electrons.
- D) CaO (Calcium oxide): This is an ionic compound. Calcium (Ca) is a metal, and oxygen (O) is a nonmetal. Calcium transfers electrons to oxygen, forming Ca2+ and O2- ions, which are strongly attracted to each other.
15. Stoichiometry: This question involves stoichiometry, which is the calculation of quantitative relationships of the reactants and products in chemical reactions. The balanced equation is:
N2 (g) + 3H2 (g) \rightarrow 2NH3 (g)
This equation tells us that 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia (NH3).
- First, find the number of moles of H_2:
Moles = \frac{mass}{molar \, mass} = \frac{9.3 \, g}{2.016 \, g/mol} \approx 4.6 \, moles - According to the balanced equation, 3 moles of H2 react with 1 mole of N2. Therefore, the number of moles of N2 needed is:
Moles \, of \, N2 = \frac{Moles \, of \, H_2}{3} = \frac{4.6 \, moles}{3} \approx 1.53 \, moles
- Now, convert moles of N_2 to grams:
Grams = moles \times molar \, mass = 1.53 \, moles \times 28.02 \, g/mol \approx 43 \, g - A) 43: correct
16. Frequency of Light: The relationship between the speed of light (c), wavelength (λ), and frequency (ν) is given by the equation:
c = \lambda \nu
Where:
- c = speed of light (3.00 \times 10^{10} cm/s)
- λ = wavelength (1.23 x 10-6 cm)
- ν = frequency (s-1)
Solve for frequency (ν):
\nu = \frac{c}{\lambda}
\nu = \frac{3.00 \times 10^{10} \, cm/s}{1.23 \times 10^{-6} \, cm} \approx 2.44 \times 10^{16} \, s^{-1} - B) 2.44 \times 10^{16}: This is the correct frequency.
17. Atomic Orbitals: The quantum numbers define the properties of atomic orbitals:
- n (principal quantum number): Describes the energy level of the electron (n = 3 means the third energy level).
- l (azimuthal quantum number): Describes the shape of the orbital (l = 0 is an s orbital, l = 1 is a p orbital, l = 2 is a d orbital).
- ml (magnetic quantum number): Describes the orientation of the orbital in space (ml = 0).
Given n = 3 and l = 1, we are looking for a p orbital in the third energy level with a specific orientation (ml = 0). P orbitals have a dumbbell shape, and since ml = 0, it should be oriented along one of the axes.
-The correct sketch (needs to be visualized) should represent a p orbital (dumbbell shape) oriented along one of the axes in a 3D space.
18. Visible Light Spectrum: The visible light spectrum ranges from violet to red. The wavelengths increase in the order:
- Violet
- Blue
- Green
- Yellow
- Orange
- Red
- C) Red: Red light has the longest wavelength in the visible spectrum.
19. Heisenberg Uncertainty Principle: The Heisenberg Uncertainty Principle states that it is impossible to determine with perfect accuracy both the position and momentum (which is mass times velocity) of a particle. The more accurately you know one, the less accurately you can know the other.
- C) momentum: The correct answer is momentum.
20. Energy of a Photon: The energy (E) of a photon is related to its wavelength (λ) by the equation:
E = \frac{hc}{\lambda}
Where:
- E = energy (5.25 \times 10^{-19} J)
- h = Planck’s constant (6.626 \times 10^{-34} J s)
- c = speed of light (3.00 \times 10^{8} m/s)
Solve for λ:
\lambda = \frac{hc}{E}
\lambda = \frac{(6.626 \times 10^{-34} \, J \cdot s)(3.00 \times 10^{8} \, m/s)}{5.25 \times 10^{-19} \, J}
\lambda \approx 3.79 \times 10^{-7} \, m - A) 3.79 \times 10^{-7}: The correct wavelength is 3.79 \times 10^{-7} m.
21. Orbitals in the Second Shell:
- The second electron shell (n = 2) has two subshells: l = 0 (s orbital) and l = 1 (p orbital).
- For l = 0 (s orbital), there is one orbital (ml = 0).
- For l = 1 (p orbital), there are three orbitals (ml = -1, 0, +1).
- Total number of orbitals in the second shell = 1 (s) + 3 (p) = 4.
- B) 4: There are 4 orbitals in the second shell.