Vegetative propagation is a form of asexual reproduction of a plant. Only one plant is involved and the offspring is the result of one parent. The new plant is genetically identical to the parent. i hope this helped <3
Resources found in lithosphere: gold and iron etc
Resources found in atmosphere: Water vapor, gases etc.
Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

Here,
is mole fraction of A,
is mole fraction of B,
is partial pressure of A and
is partial pressure of B.
The mole fraction of A and B are related to each other as follows:

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.
Now, vapor pressure can be calculated as follows:

Putting the values,

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.
Answer:

Explanation:
Hello there!
In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

Best regards!