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3 years ago

How would a bohr model of a neon atom differ from the model of nitrogen?

Chemistry

2 answers:

SVEN [57.7K]3 years ago

7 0

Answer: The Neon Atom will differ

Explanation: It will differ because the outermost energy level is not full therefor it will differ

natita [175]3 years ago

4 0

Explanation:

The Bohr hypothesis shows that an atom has an circular orbit of electrons outside the nucleus in a revolving orbit. In Bohr model for Nitrogen there are two orbits s and p. The s has two electron and p has five. Whereas in Bohr neon model there are also orbital s, and p.. The s orbital has 2 electron and the p orbital has 8 electron in it.

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Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1

timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺ + 1en → [Ni(H₂O)₄(en)]²⁺ ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0

3 years ago

Which lists the substances from least to most acidic?

Rus_ich [418]

Answer:

A. Have an amazing day!!

Explanation:

7 0

3 years ago

How many molecules are there in 840 grams of na2SO4

beks73 [17]

Answer:

3.56x10^24 molecules

Explanation:

The following data were obtained from the question:

Mass of Na2SO4 = 840g

Molar Mass of Na2SO4 = (23 x2) + 32 + (16x4) =46 + 12 + 64 = 142g/mol

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This also gives that 1mole of Na2SO4 contains 6.02x10^23 molecules.

If 1 mole (i.e 142g) contains 6.02x10^23 molecules,

Therefore, 840g of Na2SO4 will contain = (840 x 6.02x10^23)/142 = 3.56x10^24 molecules

4 0

3 years ago

Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O

Feliz [49]

Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺

3. Separate the skeleton equation into two half-reactions.

PbO₂ ⟶ Pb²⁺

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

PbO₂ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}$

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

7 0

3 years ago

A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters

Murrr4er [49]

Answer:

Explanation:

A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters

PV =nRT

at STP P= 1atm. T= 273 K

n is the number of moles. O2 has a molar mass of 32.

30 gm of O2 is 30/32= 0.94 =n

PV = nRT

at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K

1 X V = 0.94 X 0.082 X 273

using high school freshman algebra,

V= 0.94 X 0.082 X 273 = 21L

using high school algebra I,

6 0

2 years ago