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jeyben [28]
3 years ago
9

What is the energy of an electron in the third energy level of hydrogen?

Chemistry
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy.

...

Energy Levels of Electrons.

Energy Level Energy

1 -13.6 eV

2 -3.4 eV

3 -1.51 eV

4 -.85 eV

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Complete each sentence using the drop-down menus.
kirill115 [55]

Answer:

Desertification mainly impacts land  resources.

Point and nonpoint sources refer to types of water pollution.

The process of people moving to cities, called   urbanization , greatly impacts land, air, and water resources.

The burning of fossil fuels can cause smog and acid rain.

Explanation:

4 0
3 years ago
Read 2 more answers
A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? T
taurus [48]

Answer:

<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

Explanation:

  • Adding solute to water causes the depression of the freezing point.

  • We have the relation:

<em>ΔTf = Kf.m,</em>

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>Molality is the no. of moles of solute per kg of the solution.</em>

  • <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>

<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>

6 0
3 years ago
During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox
skelet666 [1.2K]

Answer:

2.9 grams.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>

5 0
3 years ago
Would the amount of solid magnesium increase decrease or stay the same if you added a strong acid to it?
Vlada [557]
Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

so the amount of solid magnesium decrease  by addition of strong acid to it.
8 0
3 years ago
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam
nikklg [1K]
The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

             CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I                0.11                       0             0
C               -x                          +x           +x
E            0.11 - x                     x             x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>


4 0
3 years ago
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