To solve this problem, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

However, since the pressure and temperature are constant, we can simplify this equation to V1/n1 = V2/n2, where V1 and n1 are the volume and number of moles of the first balloon, and V2 and n2 are the volume and number of moles of the second balloon.

Given that V1 = 1.3 L, n1 = 0.14 mol, and V2 = 7.2 L, we can substitute these values into the equation to find n2:

1.3 L / 0.14 mol = 7.2 L / n2

Solving for n2 gives:

n2 = 7.2 L * 0.14 mol / 1.3 L = 0.775 mol

So, the second balloon contains 0.775 mol of gas.

Question

To solve this problem, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

However, since the pressure and temperature are constant, we can simplify this equation to V1/n1 = V2/n2, where V1 and n1 are the volume and number of moles of the first balloon, and V2 and n2 are the volume and number of moles of the second balloon.

Given that V1 = 1.3 L, n1 = 0.14 mol, and V2 = 7.2 L, we can substitute these values into the equation to find n2:

1.3 L / 0.14 mol = 7.2 L / n2

Solving for n2 gives:

n2 = 7.2 L * 0.14 mol / 1.3 L = 0.775 mol

So, the second balloon contains 0.775 mol of gas.

Knowee AI · Accepted Answer

To solve this problem, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

However, since the pressure and temperature are constant, we can simplify this equation to V1/n1 = V2/n2, where V1 and n1 are the volume and number of moles of the first balloon, and V2 and n2 are the volume and number of moles of the second balloon.

Given that V1 = 1.3 L, n1 = 0.14 mol, and V2 = 7.2 L, we can substitute these values into the equation to find n2:

1.3 L / 0.14 mol = 7.2 L / n2

Solving for n2 gives:

n2 = 7.2 L * 0.14 mol / 1.3 L = 0.775 mol

So, the second balloon contains 0.775 mol of gas.

A balloon is inflated to a volume of 1.3 L and is found to contain 0.14 mol of a gas. If another balloon is inflated to 7.2 L at the same pressure and temperature, how many moles of gas are present in the second balloon?Multiple choice question.0.775 mol0.0253 mol1.29 mol0.826 mol

Question

Solution

Similar Questions

Upgrade your grade with Knowee