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3 years ago

Who did this

Chemistry

1 answer:

Kipish [7]3 years ago

3 0

Answer:

Niels Bohr

Explanation:

It's been long thought that the atomic model resembles a planet model, that is, it was thought that the nucleus containing neutrons and protons is the central part of the atom and there were electrons in several shells orbiting the nucleus.

This is called a planet model, since it follows the pattern seen in our galaxy: the Sun corresponds to a nucleus and the planets orbiting it are like electrons.

This model was proposed by Niels Bohr in 1915 and is called the Bohr Model. In the end, it was proved that the model wasn't really accurate, since electrons didn't have fixed orbitals but rather probabilities to be found in those orbitals, as well as the fact that electrons were exhibiting the properties of waves.

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Which of these is NOT a physical change?

alexdok [17]

Answer:

C. Magnetization of iron

Explanation:

This is a chemical change and cannot be undone by physical means therefore it isn't a physical change

6 0

3 years ago

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

Answer: The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

3 years ago

I need the answers any help can be good

aleksandrvk [35]

2.a)R b)R c)L d)L e)R f)L g)L

8 0

3 years ago

A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o

Aneli [31]

As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D

1.44 D = 4.80 × 10⁻³⁰ C.m

Dipole Moment is given as,

Dipole Moment = q  × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment =  4.80 × 10⁻³⁰ C.m

r = 163 pm = 1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

q = 2.94 × 10⁻²⁰ C

As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻

Solving for X,

X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

= 0.183 e⁻

Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.

4 0

3 years ago

To the nearest 0.5 mL, what is the water volume in the graduated cylinder? mL?

AnnZ [28]

Answer:

14cm^3

Explanation:

6 0

3 years ago