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OlgaM077 [116]
4 years ago
9

Why is it harder to remove an electron from fluorine than from carbon?

Chemistry
1 answer:
tatuchka [14]4 years ago
6 0
<span>The valence electrons of both fluorine and carbon are found at about the same distance from their respective nuclei but the greater positive charge of the fluorine nucleus attracts its valence electrons more strongly. </span>
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Please someone answer this for me ASAP
Alex777 [14]

Answer:

I'm not 100% sure but I think it is b I'm not great with math tho

6 0
3 years ago
An atom in the ground state has seven valence electrons. This atom could be an atom of which element?
Igoryamba
The atom that has 7 valence electrons is A. fluorine.
It is in the halogens group. 
It needs 1 more electron to become stable, therefore, it is highly reactive. 
4 0
4 years ago
Read 2 more answers
A mixture of helium and nitrogen gases, in a 7.03 L flask at 17 °C, contains 0.738 grams of helium and 8.98 grams of nitrogen. T
Andrei [34K]

Answer:

The partial pressure of nitrogen in the flask is 1.08 atm and the total pressure in the flask is 1.70 atm.

Explanation:

We must use the Ideal Gas Law to solve this:

Pressure . volume = n . R . T

T = T° in K → T°C + 273

17°C + 273 = 290K

n = moles

In a mixture, n is the total moles (Sum of each mol, from each gas)

Moles = Mass / Molar mass

Moles He = 0.738 g /  4g/m = 0.184 moles

Moles N₂ = 8.98 g / 28g/m = 0.320 moles

0.184 m + 0.320m = 0.504 moles

P . 7.03L = 0.504m . 0.082L.atm/ mol.K . 290K

P = (0.504m . 0.082L.atm/ mol.K . 290K) /7.03L

P = 1.70 atm  - This is the total pressure.

To know the partial pressure of N₂ we can apply, the molar fraction:

Moles of N₂ / Total moles = Pressure N₂ / Total pressure

0.320m / 0.504m = Pressure N₂ / 1.70atm

(0.320m / 0.504m) . 1.70atm = Pressure N₂

1.08atm = Pressure N₂

8 0
4 years ago
what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1
Lostsunrise [7]

Answer:

(<em>n</em> = 7) ⟶ (<em>n</em> = 4)

Explanation:

1. Convert the energy to <em>joules per mole of electrons</em>.  

<em>E</em> = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to <em>joules per electron </em>

<em>E</em> = 55 100/(6.022 × 10²³)

<em>E</em> = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to <em>calculate the transition </em>

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes  

\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})

where  

R_{\text{H}} = the Rydberg constant = 2.178 × 10⁻¹⁸ J

n_{i} and n_{f} are the initial and final energy levels.

9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})      

\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}

\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}

n_{f}^{2} = \frac{1 }{ \text{0.062 42}}

n_{f}^{2} = 16.02

n_{f} = \sqrt{16.02}

n_{f} = 4.003 \approx 4

4 0
3 years ago
For each of the following species: C2+ &amp; O2- Write the molecular orbital energy diagram and fill in the electrons Hint: in e
Tju [1.3M]

Molecular orbital energy is the energy associated with each electron in an atom or molecule.

It is expressed in electron volts (eV) and is determined by the electron's position in the atom or molecule. The molecular orbital energy diagram and fill-in the electrons are given here in each case, the number of valence electrons in the species is determined first; this is followed by the valence molecular orbital diagram for each species.

C2+: Molecular Orbital Energy Diagram

1s2 2s2 2p2

σ2s*  ← 0 e-  

σ2s   ← 2 e-

σ2p*  ← 0 e-  

σ2p   ← 0 e-

π2p*  ← 0 e-  

π2p   ← 0 e-

Bond Order: 0

Stability: Unstable

Magnetism: Diamagnetic (no unpaired electrons)

O2-: Molecular Orbital Energy Diagram

1s2 2s2 2p4

σ2s*  ← 0 e-  

σ2s   ← 2 e-

σ2p*  ← 0 e-  

σ2p   ← 2 e-

π2p*  ← 0 e-  

π2p   ← 2 e-

Bond Order: 1

Stability: Stable

Magnetism: Paramagnetic (2 unpaired electrons)

For more questions like Molecular orbital theory click the link below:

brainly.com/question/20436223

#SPJ4

6 0
2 years ago
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