In Ernest Rutherford's gold foil experiments, some alpha particles were deflected from their original paths, but most passed through the foil with no deflection. Which statement about gold atoms is supported by these experimental observations?Gold atoms consist mostly of empty space.

Gold atoms are similar to alpha particles.

Alpha particles and gold nuclei have opposite charges.

Alpha particles are less dense than gold atoms.

Answers

Answer 1

Answer:

According to ErnestRutherford's gold foil tests, the true statement is that space makes up the majority of gold atoms. The correct option is A.

According to Rutherford's studies, the majority of the alphaparticles encountered free space inside the atom since they mostly went through the gold foil with little to no deflection.

This finding helped scientists comprehend that atoms have a small, compact nucleus at their center and are primarily made up of empty space elsewhere in the atom.

The nuclearmodel of the atom was created as a result of the deflection of several alpha particles, which showed that they came into contact with the positively charged atom's nucleus.

Thus, the correct option is A.

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Your question seems incomplete, the probable complete question is:

In Ernest Rutherford's gold foil experiments, some alpha particles were deflected from their original paths, but most passed through the foil with no deflection. Which statement about gold atoms is supported by these experimental observations?

A. Gold atoms consist mostly of empty space.

B. Gold atoms are similar to alpha particles.

C. Alpha particles and gold nuclei have opposite charges.

D. Alpha particles are less dense than gold atoms

Answer 2

Answer:

Answer:

Explanation:

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Why is gas production not recognized as nitrate reduction?

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CAN SOMEONE THATS GOOD AT CHEM PLEASE HELP ME?!?!I’LL GIVE 50 POINTS!!!
AND A BRAINIEST!!!

PLEASE I HAVE OTHER HW TO DO:(

Answers

Answer:

Explanation:

1. q=mcat

q = (20)(4.184)( 30-20)

q = 836.8

Define the acid rain for class 10 .???

Answers

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Explanation:

A gas Syringe contains 42.3 milliliters of a gas at 98.15°C. Determine the volume that the gas will occupy if the temperature is decreased to -18.50°C.

Answers

The volume that the gas will occupy is "29.00 mL".

According to the question,

Volume,

$V_1 = 42.3 \ mL$
$V_2 = ?$

Temperature,

$T_1 = 98.15^(\circ) C$

$= 371.15 \ K$

$T_2 = -18.50^(\circ) C$

$= 254.5 \ K$

As we know the relation,

→ $(V_1)/(T_1) = (V_2)/(T_2)$

or,

→ $V_2 = (V_1* T_2)/(T_1)$

By substituting the values, we get

→ $= (41.3* 254.5)/(371.15)$

→ $= (10765.35)/(371.15)$

→ $= 29.00 \ mL$

Thus the above answer is right.

Learn more:

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V₁ = 42.3 mL

T₁ ( K ) = 98.15ºC + 273 = 371.15 K

V₂ = ?

T₂ ( K) = -18.50ºC + 273 = 254.5 K

V₁ / T₁ = V₂ / T₂

42.3 / 371.15 = V₂ / 254.5

371.15 x V₂ = 42.3 x 254.5

371.15 x V₂ = 10765.35

V₂ = 10765.35 / 371.15

V₂ = 29.00 mL

hope this helps!

A laboratory requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M stock solution. How could you prepare the solution needed for the lab experiment? show all work to find your answer

Answers

Answer: The volume of stock HCl solution required to make laboratory solution will be 0.25L

Explanation:

To calculate the volume of stock solution required to make the laboratory solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of laboratory solution.

We are given:

$M_1=12M\nV_1=?L\nM_2=1.5M\nV_2=2L$

Putting values in above equation, we get:

$12* V_1=1.5* 2\n\nV_1=0.25L$

Hence, the volume of stock HCl solution required to make laboratory solution will be 0.25L

The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.

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