Answer :
Part 1: 4.93 moles of contains 9.86 moles of oxygen atoms.
Part 2: 2.01 moles of contains 2.01 moles of oxygen atoms.
Explanation :
Part 1: 4.93 mol
In 1 mole of , there are 2 atoms of hydrogen and 2 atoms of oxygen.
As, 1 mole of contains 2 moles of oxygen atoms.
So, 4.93 moles of contains moles of oxygen atoms.
Thus, 4.93 moles of contains 9.86 moles of oxygen atoms.
Part 2: 2.01 mol
In 1 mole of , there are 2 atoms of nitrogen and 1 atom of oxygen.
As, 1 mole of contains 1 mole of oxygen atoms.
So, 2.01 moles of contains moles of oxygen atoms.
Thus, 2.01 moles of contains 2.01 moles of oxygen atoms.
Answer:
Covalent bond or common bond is one of the types of chemical bonds. This connection arises from electronic participation. In fact, atoms that need to receive electrons to achieve stable electron arrangement (noble gas electron arrangement or octagonal arrangement) share electrons in their valence layer with other atoms. In this case, the transfer of electrons from one atom to another does not take place, but only a pair of electrons, called a bonded or shared electron pair, belongs to the nucleus of two atoms.
Answer:
4572cm
Explanation: 1ft =30.48cm so you multiply the length value by 30.48
Answer:
HOPE IT HELPS
Complete question is;
What is the frequency of light emitted when the electron in a hydrogen atom undergoes a transition from energy level n=6 to level n=3?
Answer:
Frequency = 2.742 × 10^(14) s^(-1)
Explanation:
First of all, the energy of hydrogen electron from online values is;
E_n = -2.18 × 10^(-18) × (1/n²) J
n is the principal quantum number
We are told that hydrogen atom undergoes a transition from energy levels n = 3 to n = 6.
Thus, it means we have to find the difference between the electrons energy in the energy levels n = 3 to n = 6.
Thus;
E_n = E_6 - E_3
Thus;
E_n = [-2.18 × 10^(-18) × (1/6²)] - [-2.18 × 10^(-18) × (1/3²)]
E_n = (2.18 × 10^(-18)) × [-1/36 + 1/9]
E_n = 0.1817 × 10^(-18) J
From Planck expression, we can find the frequency. Thus;
E = hf
Where h is Planck's constant = 6.626 × 10^(-34) m²kg/s
Thus;
0.1817 × 10^(-18) = 6.626 × 10^(-34) × f
f = (0.1817 × 10^(-18))/(6.626 × 10^(-34))
f = 2.742 × 10^(14) s^(-1)
The frequency of light emitted during an electron transition in a hydrogen atom is determined by calculating the energy difference between the two energy levels and then using this to calculate the frequency using the equation for energy of a photon.
The frequency of light emitted during a transition of an electron in a hydrogen atom can be calculated using the formula for the energy difference (∆E) between two energy levels n1 and n2 in the hydrogen energy level diagram.
The formula to calculate energy difference is: ∆E = E(n2) - E(n1) where E(n) represents the energy of an energy level n. The energy difference ∆E is negative when an electron goes down an energy level (i.e., emits a photon), as the energy level n1 is greater than n2.
The frequency of the emitted photon (∆E) is then given by the formula ∆E = hf where h is Planck's constant (6.63 x 10^-34 Joule seconds) and f is the frequency. Therefore, you can rearrange the equation to find the frequency: f = ∆E / h.
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Answer:
Non polar covlant
Explanation:
Answer:
Mass = 42.6 g
Explanation:
Given data:
Mass of CF₂Cl₂ = 31.2 g
Mass of Cl₂ = ?
Solution:
Number of moles of CF₂Cl₂ = mass/molar mass
Number of moles = 31.2 g/121 gmol
Number of moles = 0.3 mol
1 mole of CF₂Cl₂ contain 2 moles of Cl atom.
0.3 mol × 2 = 0.6 mol
Mass of Cl₂:
Mass = number of moles × molar mass
Mass = 0.6 mol × 71 g/mol
Mass = 42.6 g