Observe the equation presented below.

Question

Au2O3Au+_02

To adhere to the law of conservation of mass, what set of coefficients must this equation possess, arranged from left to right?

(1 point)

2,4,3

2,1,3

1,1,3

2.4.3

Answer 1

To balance the chemical equation for the reaction of gold(III) oxide (Au₂O₃) decomposing into gold (Au) and oxygen (O₂), we need to ensure that the number of atoms of each element is equal on both sides of the equation.

The unbalanced equation is:

$${\text{Au}}_2{\text{O}}_3\to \text{Au}+{\text{O}}_2$$

Step 1: Count the atoms on the reactant side:

• There are 2 gold (Au) atoms.
• There are 3 oxygen (O) atoms.

Step 2: Count the atoms on the product side:

• Initially, there is 1 gold (Au) atom.
• There are 2 oxygen (O) atoms in O₂.

Step 3: Balance the gold atoms: To balance the gold, we will put a coefficient of 2 in front of Au on the product side:

$${\text{Au}}_2{\text{O}}_3\to 2\text{Au}+{\text{O}}_2$$

Step 4: Update the counts:

• Now there are 2 gold (Au) atoms on both sides.
• Oxygen atoms: the reactant side still has 3, and the product side has 2 from O₂.

Step 5: Balance the oxygen: To balance the oxygen, we can place a coefficient of 3/2 in front of O₂, or equivalently, we can use whole number coefficients by multiplying the entire equation by 2.

$$2{\text{Au}}_2{\text{O}}_3\to 4\text{Au}+3{\text{O}}_2$$

Now, let's count the atoms again:

• Reactants: 4 Au and 6 O (from 2 Au₂O₃).
• Products: 4 Au and 6 O (from 3 O₂).

Both sides are now balanced.

Thus, the coefficients from left to right are:

2, 4, 3.

So the answer is 2, 4, 3.

Answer 2

Chemical equations should be balanced to illustrate the law of mass conservation. Which of the following statements concerning balanced chemical equations is accurate? (1 point)

The products must have an identical arrangement of atoms as that present in the reactants.

The products must have an identical quantity of atoms but may differ in the types of atoms

compared to the reactants.

The products should possess the same quantity and variety of atoms as the reactants.

The products must have an identical variety of atoms but may vary in the quantity of atoms compared to the reactants

Answer 3

The accurate statement concerning balanced chemical equations is:

The products should possess the same quantity and variety of atoms as the reactants.

This statement aligns with the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Therefore, in a balanced chemical equation, the total number of each type of atom must be the same on both the reactant and product sides, ensuring that the variety (types of atoms) as well as the quantity of those atoms remains constant.

Answer 4

Look at the list below.

1. The division of the nucleus into fragments.

II. The combining of atomic nuclei.

III. The kind of reaction that fuels the stars.

IV. The kind of reaction that emits additional protons.

Which among these pertain to nuclear fusion reactions?

(1 point)

I and III

OII and IV

I and IV

II and III

Answer 5

Nuclear fusion reactions are characterized by the combining of atomic nuclei to form a heavier nucleus, which is accompanied by the release of energy. This process is responsible for the energy produced in stars.

Now, let's evaluate the statements:

1. The division of the nucleus into fragments. - This pertains to nuclear fission, not fusion.

II. The combining of atomic nuclei. - This describes nuclear fusion.

III. The kind of reaction that fuels the stars. - This refers to nuclear fusion, as it is the process that powers stars.

IV. The kind of reaction that emits additional protons. - This is not specific to fusion and can be misleading as it can describe different types of reactions.

Based on this evaluation, the statements that pertain to nuclear fusion reactions are:

II and III.

So, the correct answer is II and III.

Answer 6

What is the half-life of argon-39 if a sample initially weighing 1578 grams has reduced to 394.5 grams after 538 years? (1 point)

135 years

180 years

269 years

538 years

Answer 7

To determine the half-life of argon-39 based on the provided information, we can use the half-life formula, which relates the remaining amount of a substance to its initial amount and the number of half-lives that have passed.

The formula for remaining mass after a certain number of half-lives is:

$$m=m_0{\left(\frac{1}{2}\right)}^n$$

where:

• $m$ is the remaining mass,
• $m_0$ is the initial mass,
• $n$ is the number of half-lives.

In this case:

• Initial mass $m_0=1578$ grams
• Remaining mass $m=394.5$ grams
• Time elapsed $t=538$ years

First, we can determine how many half-lives have passed by rearranging the equation:

$$394.5=1578{\left(\frac{1}{2}\right)}^n$$

Dividing both sides by 1578 gives:

$${\left(\frac{1}{2}\right)}^n=\frac{394.5}{1578}$$

Calculating the right-hand side:

$$\frac{394.5}{1578}=0.25$$

We recognize that $0.25={\left(\frac{1}{2}\right)}^2$. Therefore, $n=2$.

This indicates that 2 half-lives have passed in the 538 years.

To find the half-life (denote it as $t_{1/2}$), we use:

$$2\times t_{1/2}=538\text{ years}$$

Solving for $t_{1/2}$:

$$t_{1/2}=\frac{538\text{ years}}{2}=269\text{ years}$$

Thus, the half-life of argon-39 is 269 years.

Answer 8

If only one-eighth of the sodium-24 remains, roughly how many hours have elapsed, given its half-life is approximately 15 hours? (1 point) -

30 hours

45 hours

15 hours

60 hours

Answer 9

To determine how many hours have elapsed when only one-eighth of sodium-24 remains, we can use the concept of half-lives.

If the half-life of sodium-24 is approximately 15 hours, we can observe how many half-lives it takes for the remaining amount to reduce to one-eighth.

1. After 1 half-life:

   • Remaining: $\frac{1}{2}$

2. After 2 half-lives:

   • Remaining: $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

3. After 3 half-lives:

   • Remaining: $\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

This means that it takes 3 half-lives to go from the initial amount to one-eighth of the remaining amount.

Now, we can calculate the total time that has passed:

$$\text{Total time}=3\times \text{half-life}=3\times 15\text{ hours}=45\text{ hours}$$

Therefore, if only one-eighth of sodium-24 remains, roughly 45 hours have elapsed.

Answer 10

Plutonium-239 is a radioactive isotope frequently employed as nuclear reactor fuel, with a half- ife of 24,100 years. Approximately how much time would be required for 504 grams of Plutonium-239 to undergo decay, leaving only 63 grams of plutonium-239 behind? (1 point)

96,400 years

72,300 years

24,100 years

48,200 years

Answer 11

To determine the amount of time required for 504 grams of plutonium-239 to decay to 63 grams, we will first find out how many half-lives are needed to go from 504 grams to 63 grams.

1. Find out the factor of decay: $$\text{Factor}=\frac{\text{Initial amount}}{\text{Remaining amount}}=\frac{504\text{ grams}}{63\text{ grams}}=8$$

2. Determine how many half-lives correspond to this factor: Each half-life reduces the quantity by half:

   • After 1 half-life: $\frac{504}{2}=252$ grams
   • After 2 half-lives: $\frac{252}{2}=126$ grams
   • After 3 half-lives: $\frac{126}{2}=63$ grams

   So it takes 3 half-lives to go from 504 grams to 63 grams.

3. Calculate the total time: The half-life of plutonium-239 is 24,100 years. Therefore, the time for 3 half-lives is: $$\text{Total time}=3\times 24,100\text{ years}=72,300\text{ years}$$

Thus, approximately 72,300 years would be required for 504 grams of plutonium-239 to decay to 63 grams.