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3 years ago

When Rutherford discovered the atomic nucleus it helped develop which of the following ideas about the atom?

Chemistry

2 answers:

Snowcat [4.5K]3 years ago

8 0

Answer:

The correct answer is option D.

Explanation:

Rutherford gave an experiment known as gold foil experiment.

In his experiment, he took a gold foil and bombarded it with alpha particles (carrying positive charge). He thought that the particles will pass straight through the foil, but to his surprise, many of them passed through, some of them deflected their path and a few of them bounced back.

From this he concluded that in an atom, there exist a small positive charge in the center. Due to this positive charge, the alpha particles deflected their path and some of them bounced straight back their path.

Rutherford in his model of an atom proposed that in an atom positive charge is concentrated in the center of an atom.
The center where positive charge is located was termed as nucleus by him.

MatroZZZ [7]3 years ago

7 0

The answer to this is <span>D. The nucleus contained positively-charged particles. i am positive i just finished my chemistry class</span>

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Nataly [62]

Answer:

Chemistry

Explanation:

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3 years ago

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WARRIOR [948]

Answer:

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2 years ago

At stp,a certain mass of gas occupies a volume of 790cm3. find the temperature at which the gas occupies 1000cm3 amd has a press

Luden [163]

The temperature : 332.75 K

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

volume of gas = 790 cm³ = 0.79 L

mol of gas at STP :

$\tt \dfrac{0.79}{22.4}=0.035$

Use ideal gas :

$\tt PV=nRT$

P=726 mmHg=0.955 atm

V= 1000 cm³ = 1 L

$\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{0.955\times 1}{0.035\times 0.082}\\\\T=332.75~K$

3 0

3 years ago

For the reaction below, Kp = 1.59 at 100°C. If 1.0 g of SrCO3 is placed in an empty 5.00 L reactor and allowed to reach equilibr

Sergio [31]

Answer:

$p_{CO_2}^{eq}=1.59atm$

Explanation:

Hello, in this case, one could consider the undergoing chemical reaction as:

$SrCO_3(s)\rightleftharpoons SrO(s)+CO_2(g)$

Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:

$Kp=p_{CO_2}^{eq}=1.59atm$