Answer :
- The problem is based on the proportionality between the amount of gas and its volume.
- Set up the equation: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$.
- Solve for $n_2$: $n_2 = \frac{n_1 Imes V_2}{V_1}$.
- Substitute the values and calculate: $n_2 = \frac{4.5 Imes 37.1}{45.7} = 3.65$ moles. The final answer is $\boxed{3.65 {moles}}$.
### Explanation
1. Understanding the Problem
We are given the initial amount of oxygen gas and its volume, and we are asked to find the new amount of moles when the volume is decreased. We can use the proportionality relationship between the amount of gas and the volume, assuming the temperature and pressure are constant.
2. Defining Variables
Let $n_1$ be the initial amount of oxygen gas (4.5 moles) and $V_1$ be the initial volume (45.7L). Let $n_2$ be the final amount of oxygen gas (in moles) and $V_2$ be the final volume (37.1L).
3. Setting up the Equation
The relationship between the amount of gas and the volume is given by $\frac{V_1}{n_1} = \frac{V_2}{n_2}$.
4. Solving for the Unknown
We want to solve for $n_2$, so we rearrange the equation to get $n_2 = \frac{n_1 Imes V_2}{V_1}$.
5. Substituting Values
Now, we substitute the given values into the equation: $n_2 = \frac{4.5 Imes 37.1}{45.7}$.
6. Calculating the Final Answer
Calculating the value of $n_2$, we get $n_2 = \frac{4.5 Imes 37.1}{45.7} = 3.653172866520788$. Rounding to two decimal places, we have $n_2 = 3.65$ moles.
7. Final Answer
Therefore, the new amount of moles of oxygen gas is approximately 3.65 moles.
### Examples
Imagine you're inflating a balloon. If you know how much air is initially in the balloon and its size, you can predict how much air you'll need to add to reach a desired larger size, assuming the pressure and temperature stay the same. This principle is also used in various industrial processes, such as compressing gases or calculating the amount of reactants needed in a chemical reaction.
- Set up the equation: $\frac{V_1}{n_1} = \frac{V_2}{n_2}$.
- Solve for $n_2$: $n_2 = \frac{n_1 Imes V_2}{V_1}$.
- Substitute the values and calculate: $n_2 = \frac{4.5 Imes 37.1}{45.7} = 3.65$ moles. The final answer is $\boxed{3.65 {moles}}$.
### Explanation
1. Understanding the Problem
We are given the initial amount of oxygen gas and its volume, and we are asked to find the new amount of moles when the volume is decreased. We can use the proportionality relationship between the amount of gas and the volume, assuming the temperature and pressure are constant.
2. Defining Variables
Let $n_1$ be the initial amount of oxygen gas (4.5 moles) and $V_1$ be the initial volume (45.7L). Let $n_2$ be the final amount of oxygen gas (in moles) and $V_2$ be the final volume (37.1L).
3. Setting up the Equation
The relationship between the amount of gas and the volume is given by $\frac{V_1}{n_1} = \frac{V_2}{n_2}$.
4. Solving for the Unknown
We want to solve for $n_2$, so we rearrange the equation to get $n_2 = \frac{n_1 Imes V_2}{V_1}$.
5. Substituting Values
Now, we substitute the given values into the equation: $n_2 = \frac{4.5 Imes 37.1}{45.7}$.
6. Calculating the Final Answer
Calculating the value of $n_2$, we get $n_2 = \frac{4.5 Imes 37.1}{45.7} = 3.653172866520788$. Rounding to two decimal places, we have $n_2 = 3.65$ moles.
7. Final Answer
Therefore, the new amount of moles of oxygen gas is approximately 3.65 moles.
### Examples
Imagine you're inflating a balloon. If you know how much air is initially in the balloon and its size, you can predict how much air you'll need to add to reach a desired larger size, assuming the pressure and temperature stay the same. This principle is also used in various industrial processes, such as compressing gases or calculating the amount of reactants needed in a chemical reaction.
