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3 years ago

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What properties prompted Rutherford to use alpha particles and gold? (What set of experimental conditions would make it easiest

to detect differences in atomic structure?)

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

<span>Gold is a stable heavy metal that is easily formed into uniform and thin sheets of measurable property. Alpha particles are relatively massive (compared to electrons) and so deflection is less dramatic and more easily measured. This is the property that prompted Rutherford to use alpha particles and gold. </span>
The particles that are positively charged were fired at the gold foil and when it reached that goldatoms’ nuclei, which are also positively charged, the particles were repelled back

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A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What ar

Leona [35]

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$ .

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$ .

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the n-propyl alcohol = $p^{o}_1=21.0 Torr$

Partial vapor pressure of the isopropyl alcohol = $p^{o}_2=45.2 Torr$

$p=x_1\times p^{o}_1+x_2\times p^{o}_2$ (Raoult's Law)

$p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2$

$38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr$

$x_1=0.2727$

$x_2=1-0.2727=0.7273$

$x_1\& x_2$ is mole fraction in liquid phase.

Mole fraction of components in vapor phase $y_1\& y_2$

$p_1=y_1\times p$ (Dalton's law of partial pressure)

$y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}$

$y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468$

$y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}$

$y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516$

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$

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3 years ago

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algol [13]

The heat in particles travels through convection at a certain speed depending on what density a mass has.

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3 years ago

What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?

Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

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3 years ago

sample of atmospheric gas collected at an industrial site is stored in a 250 mL amber glass bottle that has a pressure of 1.02 a

Nady [450]

Answer:- New pressure is 0.942 atm.

Solution:- The volume of the glass bottle would remain constant here and the pressure will change with the temperature.

Pressure is directly proportional to the kelvin temperature. The equation used here is:

$P_1T_2=P_2T_1$

Where, $T_1$ and $T_2$ are initial and final temperatures, $P_1$ and $P_2$ are initial and final pressures.

$T_1$ = 20.3 + 273.15 = 293.45 K

$T_2$ = -2.0 + 273.15 = 271.15 K

$P_1$ = 1.02 atm

$T_2$ = ?

Let's plug in the values in the equation and solve it for final pressure.

$1.02atm(271.15K)=P_2(293.45K)$

$P_2=\frac{1.02atm*271.15K}{293.45K}$

$P_2$ = 0.942 atm

So, the new pressure of the jar is 0.942 atm.

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3 years ago