Answer:
The common hippopotamus, or hippo, is a large, mostly herbivorous, semiaquatic mammal and ungulate native to sub-Saharan Africa. It is one of only two extant species in the family Hippopotamidae, the other being the pygmy hippopotamus. The name comes from the ancient Greek for "river horse".
Answer:
Organic matter composted with earthworms, often known as vermicompost, has been demonstrated in studies to have disease-suppressing qualities. The earthworms loosen the soil and enrich it with humus. As a result, they are known as farmer's pals. They help the crops grow healthier and over the time have earned farmers trust.
Explanation:
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Answer:
Explanation:
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In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:
Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:
Then, we calculate the partial pressure as shown below:
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Answer:
compound
Explanation:
"Marble is typically more than 95% calcium carbonate, perhaps even 99% calcium carbonate, and calcium carbonate is a compound. Marble has a good claim to be recognized as a compound."
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
