When, it donates electons.
as for example take
NaCl ( sodium chloride)
it's an ionic compund,
that means it is formed by donating or gaining electrons
Na is writen first than, it must be electropositive i.e it has donated electons which made it positive and the clorine gains electron so it's electronegative.
Na is positive because
as we know it's atomic number is 11 that means it has 11 protons and 11 electrons
now, when it donate electon it has, greater number of protons whose change is +ve so the atom becomes overall positively charged ion or cation.
and something same happens in clorine and because it gains one electron and the number of electrons increase in it by 1 whise charge is -ve so, the atom becomes negatively charged ion or anion which has a -1 charge.
Halogens can be characterized as :
1) High Electronegativity
2) Low melting & boiling points
3) Forms Ionic compounds
4) Non-metallic in nature
Hope this helps!
System A undergoes an increase in entropy while system B undergoes a decrease in entropy.
Entropy is the degree of disorderliness of a system. The entropy of a system depends on the number of particles present in the system as well as the state of matter.
Entropy is increased when solid particles dissolve in water because more particles are produced thereby increasing the level of disorderliness in the system.
On the other hand, when vapor is condensed, the degree of disorderliness decreases as gases are converted to liquids.
Learn more: brainly.com/question/13146879
CIO4-=-1
CI=4O=-1
O has a 2- oxidation change so
CI+4(-2)=-1
CI+(-8)=-1
CI=-1+8=7
So the oxidation number of chlorine is 7 in this case
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>