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3 years ago

10

Statement 1: The reaction follows the mass energy equation E= mc". Statement 2: A large amount of energy is produced during the

reaction.
Which statement(s) are true for both fusion and fission reactions? (1 point)
a. Only 1
b. Only 2
c. Both 1 and 2
d. Neither 1 nor 2

1 answer:

pogonyaev3 years ago

4 0

Answer:

d

Explanation:

Chem teacher

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What type of electromagnetic waves does the human eye see as the colors red, blue, or green

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Helium is the lightest noble gas and the second most abundant element (after hydrogen) in the universe. The mass of a helium−4 a

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<u>Answer:</u> The fraction of atom's mass contributed by nucleus is 0.99

<u>Explanation:</u>

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

The isotopic symbol of Helium-4 atom is $_2^4\textrm{He}$

Number of electrons = 2

Number of protons = 2

Number of neutrons = 4 - 2 = 2

We are given:

Mass of He-4 atom = $6.64648\times 10^{-24}g$

Mass of 1 electron = $9.10939\times 10^{-28}g$

Calculating the mass contributed by the nucleus = $m_{He}-2(m_e)$

Mass of the nucleus of He-4 atom = $6.64648\times 10^{-24}-(2\times (9.10939\times 10^{-28}))=(6.64648-0.0018219)\times 10^{-24}=6.64466\times 10^{-24}$

To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:

$\text{Fraction of atom's mass contributed by nucleus}=\frac{\text{Mass of nucleus}}{\text{Mass of atom}}$

Putting values in above equation, we get:

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2 years ago

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2 years ago

What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

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Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

$2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3$ ;

$4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}$ .

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

$3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}$ .

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

$2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}$

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

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