Sodium has a lower ionization energy than magnesium describes why sodium reacts vigorously than magnesium chloride.
<h3>Why is sodium more reactive than magnesium?</h3>
- Sodium is more reactive than magnesium because it has the ability to easily lose electron, hence have lower ionization energy.
- Sodium belong to group one on the periodic table and they are called akali metal while magnesium belong to group two on the periodic table and they are called alkali Earth metal.
- Sodium and magnesium belong to the in the 3rd period. Iin the outermost energy level sodium has one electron but magnesium has 2 electrons. Therefore, there is more attraction abetween the nucleus and electrons in magnesium than that of sodium.
Therefore, sodium is more reactive than magnesium chloride because of lower ionization energy.
For more details on sodium reactivity, check the link below.
brainly.com/question/6837593
Answer:
finding the mass percentage oven element in a compound might sound complicated, but the calculation is simple. For example, to determine the mass percentage of hydrogen in water H2O, divide the major mass of hydrogen by the total molar mass of water and then multiply the result by 100
Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
Answer:
Molarity is halved when the volume of solvent is doubled.
Explanation:
Using the dilution equation (volume 1)(molarity 1)=(volume 2)(molarity 2), we can demonstrate the effects of doubling volume.
Suppose the starting volume is 1 L and the starting molarity is 1 M, and doubling the volume would make the final volume 2 L.
Plugging these numbers into the equation, we can figure out the final molarity.
(1 L)(1 M)=(2 L)(X M)
X M= (1 L x 1 M)/(2 L)
X M= 1/2 M
This shows that the molarity is halved when the volume of solvent is doubled.
