1. What volume will a sample of hydrogen occupy at 128.0 oC if the gas occupies a volume of 2.23 dm3 at a temperature of 0.0 oC? Assume that the pressure remains constant. (remember to change to Kelvin).

2. If a gas occupies 1733 cm3 at 10.0 oC, at what temperature will it occupy 950 cm3? Assume that pressure remains constant.

3. A gas occupies 1560 cm3 at 285 K. To what temperature must the gas be lowered to, if it is to occupy 25.0 cm3? Assume a constant pressure

1. (V1/T1) = (V2/T2)

2. Same formula.
3. same formula.

1. To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature, assuming pressure remains constant. Charles's Law can be represented by the equation: V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Given:
V1 = 2.23 dm^3
T1 = 0.0 oC (convert to Kelvin by adding 273.15 -> 273.15 + 0.0 = 273.15 K)
T2 = 128.0 oC (convert to Kelvin -> 273.15 + 128.0 = 401.15 K)

Using the Charles's Law equation, we can rearrange it to solve for V2:
V2 = (V1 * T2) / T1

Substituting the given values:
V2 = (2.23 dm^3 * 401.15 K) / 273.15 K

Now, we need to convert the answer back to cm^3 since the given volume is in cm^3:
V2 = (2.23 dm^3 * 401.15 K) / 273.15 K * (1000 cm^3 / 1 dm^3)
V2 = 3262 cm^3 (rounded to the nearest whole number)

Therefore, the volume of the sample of hydrogen at 128.0 oC would be 3262 cm^3.

2. In this problem, we need to use Charles's Law again to find the final temperature. Using the same equation as above (V1/T1 = V2/T2):

Given:
V1 = 1733 cm^3
T1 = 10.0 oC (convert to Kelvin -> 273.15 + 10.0 = 283.15 K)
V2 = 950 cm^3
T2 = unknown (to be solved)

Rearranging the equation to solve for T2:
T2 = (T1 * V2) / V1

Substituting the given values:
T2 = (283.15 K * 950 cm^3) / 1733 cm^3

Therefore, the temperature at which the gas will occupy 950 cm^3 is approximately 155.35 K.

3. This problem involves the same concept, Charles's Law, to find the final temperature. Rearranging the equation V1/T1 = V2/T2:

Given:
V1 = 1560 cm^3
T1 = 285 K
V2 = 25.0 cm^3
T2 = unknown (to be solved)

Solving for T2:
T2 = (T1 * V2) / V1

Substituting the given values:
T2 = (285 K * 25.0 cm^3) / 1560 cm^3

Therefore, the gas must be cooled to approximately 4.60 K for it to occupy 25.0 cm^3.