Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.
Answer: 43 L
They differ in their molecular structures and properties.
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
The original concentration of the acid solution is 6.175 
10^-4 mol / L.
<u>Explanation:</u>
Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume
HBr + NaOH -----> NaBr + H2O
There is a 1:1 equivalence with acid and base.
Moles of NaOH = 72.90 10^-3 0.25
= 0.0182 mol.
[ HBr ] = moles of base / volume of a solution
= 0.0182 / 29.47
= 6.175 10^-4 mol / L.
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
