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faust18 [17]
2 years ago
10

How might this graph be different if the farmers had grown a wide variety of potato plants or had included other crops in their

fields? Construct an explanation to support your answer. Construct an explanation to support your answer.

Chemistry
1 answer:
mote1985 [20]2 years ago
3 0

The graph would not have plummeted if various species were planted due to improvement of the potato crops.

<h3>What is variety?</h3>

The term variety has to do with the planting of different species of plants on the same farm. We know that the planting of different species of the same crop helps to improve the resistance to diseases of a plant.

Hence, the graph would not have plummeted if various species were planted due to improvement of the potato crops.

Learn more about improved specie:brainly.com/question/3627477

#SPJ1

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How much energy is used to melt 44.33 g of solid oxygen?
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Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

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Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

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