Answer:
At 30 and 2,204
diagonal
liquid phase
2856
top horizontal line
flat
the change from a solid to a liquid
Explanation:
Nature is full of wonders and one of the most impressive of them are symbiotic relationships (mutually beneficial relationships). Despite the competition in the kingdom of life, here 2 actors help each other to gain more than they could individually. Let's look at the answers. The sea anemone can't possibly fertilize the clownfish's eggs, only another clownfish can do that. The 4th choice might be correct but it does not help the clownfish; The 5th choice is also correct but it is not related to the offspring of the clownfish. For the 3rd choice, we have that the anemone sometimes provides leftovers for the fish, but the fish' eggs cannot use this source of nutrients. However, we know that the 1st assertion holds; the poisonous tentacles of the sea anemone protect the eggs from potential predators.
Answer:
The reaction D has the value of ΔH°rxn equal to ΔH°f for the product.
Explanation:
The ΔH°f for product is equal to ΔH°rxn when the reagents are in their elemental state (ΔH°f = 0) and form one mole of product.
We have to find the reagents that are in their elemental state and that only form one mole of product:
A) 2Ca (s) + O₂ (g) → 2CaO (s)
The reagents are in their elemental state but the reaction forms two mole of product.
B) C₂H₂ (g) + H₂ (g) → C₂H₄ (g)
C₂H₂ (g) is not in its elemental state.
C) 2C (graphite) + O₂ (g) → 2CO (g)
Graphite and Oxygen are in their elemental state but the reaction forms two mole of product.
D) 3Mg (s) + N₂ (g) → Mg₃N₂ (s)
Magnesium and Nytrogen are in their elemental state and the reaction forms one mole of product.
E) C (diamond) + O₂ (g) → CO₂ (g)
Diamond is not in its elemental state.
Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
