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4 years ago

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Who is known as the "father of modern chemistry” because he first organized all known elements into four different groups? Johan

n Wolfgang Döbereiner Antoine Lavoisier Dmitri Mendeleev Henry Moseley

2 answers:

scoundrel [369]4 years ago

7 0

Answer Henry Moseley

Sonja [21]4 years ago

6 0

Answer:

D

Explanation:

the answer is not D

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When considering the relationship among standard free energy change, equilibrium constants, and standard cell potential, the equ

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ΔG° = - RTLnK is used to find the standard cell potential given the equilibrium constant

Explanation:

for an ideal disolution:

⇒ ΔG = RT∑ni LnXi

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∴ μ : chemical potential

∴ μ*: chem. potential of the pure component at T and P.

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4 years ago

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3 years ago

German physicist Werner Heisenberg related the uncertainty of an object\'s position (Δx) to the uncertainty in its velocity Δv.

Assoli18 [71]

Heisenberg's <em>Uncertainty Principle</em> gives a relationship between the standard deviation of an object's position and its momentum.

$\Delta p \cdot \Delta x = h / (4 \pi)$ where

$\Delta p$ the standard deviation of the object's <em>momentum,</em>
$\Delta x$ the standard deviation of the object's <em>position, </em>and
$h \approx 6.63 \times 10^{-34} \; \text{J} \cdot \text{s}$ the Planck's constant.

By definition, the momentum of the electron equals the product of its mass and velocity.

$p = m\cdot v$

Assuming that measurement of the mass of the electron $m$ is accurate. It is assumed to be a coefficient of constant value. The <em>standard deviation</em> in the electron's velocity is thus directly related to that of its mass. That is:

$\Delta p = m \cdot \Delta v$

$\Delta v = 0.01 \times 10^{6} \;\text{m}\cdot \text{s}^{-1}$ from the question;

$\Delta p = m\cdot v \\ \phantom{\Delta p} = 0.01 \times 10^{6} \; \text{m} \cdot \text{s}^{-1} \times 9.11 \times 10^{-31} \; \text{kg}\\\phantom{\Delta p} = 9.11 \times 10^{-27} \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1}$

Convert the unit of the Planck's constant to base SI units (kg, m, s, etc.) if it was provided in derived units such as joules. Doing so would allow for a dimension analysis on the accuracy of the result.

$h = 6.63 \times 10^{-34} \; \text{J} \cdot \text{s}\\\phantom{h} = 6.63 \times 10^{-34} \; (\text{N}\cdot \text{m}) \cdot \text{s} \\\phantom{h} = 6.63 \times 10^{-34} \; ((\text{kg} \cdot \text{m}\cdot \text{s}^{-2}) \cdot \text{m}) \cdot \text{s}\\\phantom{h} = 6.63 \times 10^{-34} \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1}$

Apply the <em>Uncertainty Principle</em>:

$\Delta x = h/ (4 \pi \cdot \Delta p)\\\phantom{\Delta x} = 6.63 \times 10^{-34} \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1} / (4 \pi \cdot 9.11\times 10^{-27} \; \text{kg} \cdot \text{m}\cdot \text{s}^{-1})\\\phantom{\Delta x} = 5.79 \times 10^{-9} \; \text{m}$ .

Dimensional analysis:

$\Delta x$ resembles the <em>standard deviation</em> of a position measurement. It is expected to have a unit of meter, which is the same as that of position.

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