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Anuta_ua [19.1K]
3 years ago
14

Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class de

cided to use images to feature the scientist of their choice.
Which image would Tumu most likely use in his assignment to feature Rudolf Virchow?

the magnified shapes of cells in a piece of cork
the different cells that make up a plant leaf
the types of cells at work in the human body
the way root cells reproduce to increase root length
Chemistry
2 answers:
Zarrin [17]3 years ago
7 0

Answer:

D the way root cells reproduce to increase root length

Explanation:

Have a good day!

xxMikexx [17]3 years ago
4 0

Answer:

The types of cells at work in the human body

Explanation:

Rudolf Virchow was a German physician who is known as the father of modern pathology. He contributed to the development of the cell theory by concluding that all cells come from preexisting cells. This led to the conclusion that diseases occur at the cellular level, which allowed scientists to diagnose diseases more accurately.

Considering that Virchow analyzed human cells, the image that Tumu would be most likely to use in his assignment is the one of the types of cells at work in the human body.

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Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

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Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,  

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= 64.0 grams / 32.0 grams per mole

= 2 mol

The amount of energy given by the process of condensation is,  

ΔH = 2 mol × 35.3 kJ/mol = 70.6 kJ

In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,  

W = -P ΔV

Let us first find the volume of 2.0 mole gas at 64 °C, or 64 + 273 = 337 K,  

PV = nRT

V = nRT/P

= 2 mol × 0.08206 L atm per mol K × 337 K/1 atm

= 55.3 L

As the liquid condenses in the process, the change in volume would be negligible. So, the volume change will be -55.3 L

W = - 1 atm × - 55.3 L

W = 55.3 L.atm

W = 55.3 L.atm × 101.3 J/1 L atm = 5602 J

W = 5602 × 1 kJ / 1000 J = 5.602 kJ

W = 5.602 kJ

Now U can be calculated using the formula,  

U = q + W

= -70.6 kJ + 5.602 kJ

= -65. kJ

Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.  

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