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4 years ago

A particle that orbits the nucleus in an atom is called an

2 answers:

8 0

This would be a proton.It has a positive charge and is found in the nucleus. Don't get it confused with an electron this is not inside the nucleus but floats around the nucleus.

ruslelena [56]4 years ago

4 0

Good Morning

Answer : A particle that orbits the nucleus in an atom is called an Electron

Hint : Electron, negatively charged particle orbiting the nucleus. Same number as protons in neutral atoms

I hope that's help !

Happy Sunday :)

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A charged atom or particle is a(n) _____. compound ion neutron valence

mel-nik [20]

This would be an ion

Hope this helps :)

4 0

4 years ago

Report with correct significant figures in scientific notation <br>3.698 X 10^4÷1.85 X 10^2

Alina [70]

Answer:

$2.00x10^2$

Explanation:

Hello!

In this case, we need to keep in mind that performing these types of operations require that the final result was shown with the same significant figures as the initial number with the fewest ones; it means, we show the final result with 3 significant figures, because 3.698 has four and 1.85 has three significant figures; therefore, we obtain:

$3.698 x 10^4/1.85x10^2\\\\2.00x10^2$

Best regards!

5 0

3 years ago

(a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate. (b) Calculate the pH of a buffer fo

zepelin [54]

Answer:

fe9ufeohdwbkdwsdjvdwihdwkbfw

8 0

3 years ago

A proton is fired toward a lead nucleus from very far away. How much initial kinetic energy does the proton need to reach a turn

Olegator [25]

Answer:

The electric force is conservative.

Since

ΔK = −ΔU,

Kf − Ki =Ui −Uf.

We have,

Kf = 0

Ui = 0.

Thus Ki =Uf.

6 0

4 years ago

ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion

vfiekz [6]

Answer:

Here's what I get

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

3. Set up an ICE table

A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹: 0.35 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.35-x x x

4. Solve for x

$\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

3 0

3 years ago