Answer:
The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.
Explanation:
Mg produces less amount of MgO than O2; therefore Mg is the limiting reagent. O2 produces more amount of MgO than Mg; therefore O2 is the excess reagent.
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
There are 4 significant figures! Start counting after the first non-zero digit :)
Hope this helps.
Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = 
= specific heat of liquid benzene = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
2p
Explanation:
it has 3 dumbell shapes, hence p
you can't determine the principal quantum number by looking at the shape, however bigger or spread orbital means higher value of n
