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Crank
2 years ago
12

SOME ONE PLEASE HELP ME I NEED THIS SO BAD 30 POINTS AND WIL MARK BRAINLIST!!!

Chemistry
2 answers:
Ivanshal [37]2 years ago
7 0

Answer:

The Predator is C, top of the pyramid

Secondary Consumers is E, the bottom of the pyramid

Primary Consumers is B, second level from the bottom

Producers is D, bottom of the pyramid

Tertiary Consumers is A, second level from the top

(btw i tried my best lol, i dont know much about this)

Explanation:

Phantasy [73]2 years ago
4 0

Answer:

Ape x Predator - C)

Tetiary Consumers - A)

Secondary Consumers - E)

Primary Consumers - B)

Producers - D)

Explanation:

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Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

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3 years ago
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