Answer:
Amplitude is measured from the center line to the highest point in the waves.
Wavelength is the distance between one wave to the other from the highest point.
Frequency is the rate of the waves.
Answer:
2.4 g
Explanation:
Step 1: Given data
- Initial pressure (P₁): 755 torr
- Final pressure (P₂): 1.87 atm
Step 2: Convert "P₁" to atm
We will use the conversion factor 1 atm = 760 torr.
755 torr × 1 atm/760 torr = 0.993 atm
Step 3: Convert "T" to K
We will use the following expression.
K = °C + 273.15
K = 25°C + 273.15 = 298 K
Step 4: Calculate the initial number of moles of He
We will use the ideal gas equation.
P₁ × V = n₁ × R × T
n₁ = P₁ × V/R × T
n₁ = 0.993 atm × 16.8 L/(0.0821 atm.L/mol.K) × 298 K
n₁ = 0.682 mol
Step 5: Calculate the final number of moles of He
We will use the ideal gas equation.
P₂ × V = n₂ × R × T
n₂ = P₂ × V/R × T
n₂ = 1.87 atm × 16.8 L/(0.0821 atm.L/mol.K) × 298 K
n₂ = 1.28 mol
Step 6: Calculate the moles of He added
n = n₂ - n₁
n = 1.28 mol - 0.682 mol
n = 0.60 mol
Step 7: Convert "n" to mass
The molar mass of He is 4.00 g/mol
0.60 mol × 4.00 g/mol = 2.4 g
Answer: 3p Orbitals
Explanation:
Electrons present in the 3p orbitals are farthest from the nucleus. Therefore, the electrons present in the 3p orbital will be shielded by the electrons present in the inner orbitals. Hence, 3p orbital in sulfur is most shielded from the nuclear charge".
<h3>Answer:</h3>
Rubidium (Rb)
<h3>Explanation:</h3>
Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".
<h3>Trends in Periodic table:</h3>
Along Periods:
Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.
Along Groups:
Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.
<h3>Conclusion:</h3>
Given elements belong to same group hence, Rubidium present at the bottom of remaining elements will have least ionization energy due to facts explained in trends of groups above.
