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nikklg [1K]
2 years ago
7

An electron has an uncertainty in its position of 471 pm .

Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
3 0

<u>Answer:</u>

$\Delta x . \Delta p \geq \frac{h}{4 \pi}$

∆x is the Uncertainty in the position of the particle

∆p is the Uncertainty in the momentum of the particle

$\Delta p=\frac{h}{4 \pi \times m \times \Delta x}$

$=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} s^{-1}}{4 \times 3.14 \times 9.109 \times 10^{-31} \mathrm{kg} \times 471 \times 10^{-12} \mathrm{m}}$

$=\frac{6.626 \times 10^{-34} \mathrm{ms}^{-1}}{53887 \times 10^{-43}}$

=1.23\times10^5 ms^{-1} is the answer

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Andreyy89

Answer:

<u>B</u> What role do animals have in the carbon cycle?

Group of answer choices

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4 0
3 years ago
From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s)
gregori [183]

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

<em>(1) </em>V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

<em>(2) </em>V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = <em>-135,6 kJ</em>

ΔS° = - 8,3 J/K - 0,2 J/K =<em> -8,5 J/K</em>

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

<em>ΔG° = -133,1 kJ</em>

I hope it helps!

7 0
3 years ago
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