Answer:
All of these properties are due to the chemical structure of the compound. The chemical structure includes the bonding angle, the type of bonds, the size of the molecule, and the interactions between molecules. Slight changes in the chemical structure can drastically affect the properties of the compound.
Explanation:
Hope it helps.
Please mark as brainliest.
We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, 
moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
Answer: 7
Explanation: oxygen has 5 atoms and nitrogen has 2. Put them together and you get 7
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
I believe c is the right answer.
