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nikdorinn [45]
3 years ago
12

Which of the following scientists made an important contribution to quantum mechanics: (i) Bohr (ii) deBroglie (iii) Heisenberg

(iv) Rutherford (v) Schrodinger (select all that apply)
Chemistry
1 answer:
grandymaker [24]3 years ago
4 0

Answer:

(i) Bohr; (ii) de Broglie; (iii) Heisenberg (v) Schrödinger

Explanation:

(i) Niels Bohr — 1913 — proposed that electrons travel in fixed orbits with <em>quantized energy levels</em> and that they jump from one energy level to another by absorbing or emitting quanta of light.

(ii) <em>Louis de Broglie</em> — 1924 — proposed the wave nature of electrons and suggested that all matter behaves as both waves and particles (<em>wave-particle duality</em>).

(iii) Werner Heisenberg — 1927 — formulated quantum mechanics in terms of matrices and proposed his famous <em>uncertainty principle</em>.

(v) Erwin Schrödinger — 1926 — applied wave mechanics to the electron in a hydrogen atom, showing that electrons exist in <em>orbitals </em>rather that orbits.

(iv) <em>Ernest Rutherford</em> — 1911 — proposed that atoms have most of their mass in a central nucleus (<em>nuclear atom</em>). Quantum mechanics had not yet been invented.

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What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

m= molality

molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}

given \Delta T=6.28^{\circ}C

Molar mass of solute = 46.0 gmol^{-1}

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

molality=\frac{x}{46gmol^{-1}\times0.15kg}

6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}

x = 17.12 g



3 0
3 years ago
Does the pH increase or decrease, and does it do so to a large or small extent, with each of the following additions?(c) 5 drops
Umnica [9.8K]

The pH decreases to a large or small extent with each of the given additions.

<h3>What is common name of NaOH?</h3>

The common name of NaOH is sodium hydroxide. Lye and caustic soda are other names for sodium hydroxide, an inorganic substance having the formula NaOH. It is a white, solid ionic substance made up of the cations sodium (Na+) and the anions hydroxide (OH). Sodium hydroxide is a chemical that manufacturers utilize to make things like soap, rayon, paper, explosives, colors, and petroleum products. Processing cotton fabrics, metal cleaning and processing, oxide coating, electroplating, and electrolytic extraction are further uses for sodium hydroxide. A caustic metallic base is sodium hydroxide (NaOH), sometimes referred to as lye or caustic soda. Caustic soda, an alkali, is commonly employed in a variety of sectors, primarily as a potent chemical base in the production of pulp and paper, textiles, drinking water, and detergents.

To learn more about sodium hydroxide, visit:

brainly.com/question/24010534

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6 0
1 year ago
Determine the number of moles of water assicated with the salt.
Nookie1986 [14]

Answer:

Divide the mass of your anhydrous (heated) salt sample by the molar mass of the anhydrous compound to get the number of moles of compound present. In our example, 16 grams / 160 grams per mole = 0.1 moles. Divide the mass of water lost when you heated the salt by the molar mass of water, roughly 18 grams per mole.In order to determine the formula of the hydrate, [Anhydrous Solid⋅xH2O], the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid (Equation 2.12. 6).

6 0
3 years ago
PLS HELP
alexandr1967 [171]

Answer:

Explanation:

A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1

5 0
3 years ago
Read 2 more answers
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
2 years ago
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