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MakcuM [25]
2 years ago
12

A student drew a diagram of the quantum model of an atom, as shown. A small circle is shown. Six light gray spheres and six dark

gray spheres are shown inside the circle. On the outer side of the circle is a diffused shaded circular region. This diffused shaded circular region has two distinct bands of dark gray shading in it. Which of the following explains if the student's diagram is correct or incorrect? The diagram is incorrect because electrons follow orbital rings around the nucleus. The diagram is incorrect because there is only a single type of particle at the center of the atom. The diagram is correct because electrons are present in a negatively charged cloud outside the nucleus. The diagram is correct because all electrons can be accommodated in two electron clouds around the nucleus.
Chemistry
1 answer:
Rufina [12.5K]2 years ago
8 0

Answer:

The diagram is incorrect because there is only a single type of particle at the center of the atom.

Explanation:

The small circle at the centre of the sphere which contains six light gray spheres and six dark gray spheres represents the nucleus of the atom which is found at the centre of the atom.

The nucleus contains only one type of particle which is the proton. Hence the representation of two particles at the centre of the circle makes the model incorrect.

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nekit [7.7K]

From,

RAM=element×its relative abudance/total abudance

=((107×13)+(12×109))/25

The answer is=107.96

3 0
3 years ago
How many moles are in 337 grams of tellurium?
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3 0
3 years ago
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

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Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
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xeze [42]

Answer:0.1677M

Explanation:

Molarity=moles/volume

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Once you get the number of moles, you apply it to the molarity formula.

8 0
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