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posledela
2 years ago
7

Which of the following was NOT suggested by Rutherford’s gold foil experiment?

Chemistry
1 answer:
Phoenix [80]2 years ago
5 0

Answer: The Answer is C.

Explanation: The Nucleus only makes up less than .01% of the volume of the Atom. The Nucleus does contain more than 99.9% of the mass of the Atom. I hope that this helps you! Good luck!

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what is the daughter nucleus (nuclide) produces when 64^Cu undergoes beta decay by emitting an electron? replace ? ? ?
Maurinko [17]

Answer:

Explanation: that decays by emitting an electron  particle the resulting nuclide is zinc-64. and yeah

4 0
3 years ago
The diagram shows the movement of particles from one end of the container to the opposite end of the container.
Aleksandr-060686 [28]
The correct option is this: EFFUSION BECAUSE THERE IS A MOVEMENT  OF A GAS THROUGH A SMALL OPENING INTO A LARGER VOLUME.
Effusion refers to the movement of gas particles through a small hole. According to Graham's law, the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

7 0
3 years ago
Read 2 more answers
What is the speed at which molecules or atoms move dependent on temperature and state of matter.
den301095 [7]
The hotter it gets, the faster molecules move, solid form is in low temperature, liquid in medium temperature and gas in high temperature.
3 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
HHHAAAAALLLPPPP<br><br> It’s due tomorrow
iris [78.8K]

Answer:

D.

Explanation:

Only 0.0035% of the electromagnetic spectrum is visible to the human eye

Hopefully this helped :)

7 0
3 years ago
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