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Natalija [7]
3 years ago
13

Which statement regarding the gold foil experiment is FALSE? 1. The α particles were repelled by electrons. 2. It suggested that

atoms are mostly empty space. 3. It suggested the nuclear model of the atom. 4. Most of the α particles passed through the foil undeflected. 5. It was performed by Rutherford and his research group early in the 20th century.
Chemistry
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

1. The α particles were repelled by electrons.

Explanation:

The gold foil experiment was performed by Rutherford and his research group in 1911 (at the beginning of the 20th century). In this experiment, α particles were bombed to gold foils, and films were placed surround it to collect the particles.

It was observed that most of the particles passed through of the foil undeflected, and for that, Rutherford stated that the atom was a "huge empty". Some particles were deflected, because they're attracted to the electrons at the electrosphere, and a small number of particles were complete deflected to the origin because they chocked with the small positive nuclei.

Thus, the experiment suggested the nuclear model of the atom, called the planetary model, that was improved after by Bohr and other scientists in the quantum model.

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The integrated rate law for a second-order reaction is given by:

\frac{1}{[A]t} =   \frac{1}{[A]0} + kt

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction


<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min


Hence, \frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)

<span>                                        = 4.858</span>

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>
</span>

<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

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3 years ago
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Copper is malleable, ductile, and conducts heat and electricity. Would you classify copper as a metal, nonmetal, or metalloid? W
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Answer:

The elements can be classified as metals, nonmetals, or metalloids. Metals are good conductors of heat and electricity, and are malleable (they can be ... and electricity, and are not malleable or ductile; many of the elemental nonmetals are ... under certain circumstances, several of them can be made to conduct electricity.

Hope this helps!

Explanation:

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According to the Charles' Law, the volume of a gas is proportional to temperature when pressure is constant. When going from New York to Florida, if the pressure is left constant the volume of the tires will increase.

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Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.
lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

<u>Step 1</u>: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

<u>Step 2:</u> The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

<u>Step 3: </u>Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) +  I-(aq)

<u>Step 4</u>: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

<u>Step 5:</u> Calculate molarity of iodide ion

Molarity = moles I- / volume

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Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

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