-70°C
Sink
little
hydrogen bonding
Explanation:
Completing the statements:
Water's boiling point would have been close to -70°C. Ice would sink in water. Water would release little heat to warm land during the winter. Ice is less dense than water because of the hydrogen bonding that forms a hexagonal structure in water.
The unique property of water is as a result of its hydrogen bonding. Water is a polar covalent compound. Like most covalent compound, water would have naturally had a very low boiling point.
The intermolecular forces all hydrogen bonding gives water its unique nature.
Hydrogen bond is formed by an attraction between hydrogen one water water molecule and more electronegative atom on another molecule usually oxygen, nitrogen and fluorine.
They form very strong intermolecular interaction responsible for the behavior of water.
The higher specific heat capacity of water is due to this bond. It absorbs a lot of heat and does not release them on time. This causes water release heat during winter.
Water has a hexagonal shape or structure linking each molecules.
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Hydrogen bonding brainly.com/question/10602513
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<span>To convert the number of moles a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol. In this case, we are given with 7.02 moles of sodium chloride. Hence the answer is 4.23 x 10^24 formula units.</span>
It is two or more objects and different things that can be removed from each other.
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
