Answer:
yahh
Explanation:
a precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed.
Answer:
Explanation:
Molarity = number of moles / volume
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M
Answer:
2.8 x 10²³ molecules H₂O
1.4 x 10²³ molecules O₂
Explanation:
First, you will need the balanced chemical equation for the formation of water:
2H₂ + O₂ -> 2H₂O
This will help in determining the mole ratios between water and oxygen, which we will need later.
Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules
8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O
Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O
Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂
8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂
Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂
No, xenon will not react on Nitrogen because xenon only reacts with oxygen and fluorine. Xenon only reacts with the 2 elements because these 2 elements contain a strong electronegative element which is high in energy barrier that can move the electrons inside xenon elements. Xenon is an element with a symbol of Xe, it is odorless and colorless gas. Common uses of xenon are flash lamps and arc lamps. Xenon is a luminous element, it glows when placed near to high voltage electric field.