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Lena [83]
3 years ago
7

Yurr wussup can someone help me with my questions on for science if nayones down put in the cmts . :)

Chemistry
2 answers:
barxatty [35]3 years ago
8 0

Answer:

Sure

Explanation:

Ira Lisetskai [31]3 years ago
3 0

Answer:

bet

Explanation:

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Which of the following scientists made an important contribution to quantum mechanics: (i) Bohr (ii) deBroglie (iii) Heisenberg
grandymaker [24]

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.

4 0
3 years ago
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
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What is the best definition
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B.

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thermal expansion♡

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