Answer:
B. increase the distance between the objects
Explanation:
A) increase the mass of one of the objects
Answer:
As the average kinetic energy increases, the particles move faster and collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature
HOPE THIS HELPED!!!!!!!!!!!!!! XDDDDD
One of the differences I can think of is that hydrogen is no longer listed as a group I element.
According to the mendeleev tables that I looked up, hydrogen is catorgrized as a group I element, along with Lithium, sodium, Potassium etc. However, nowadays, hydrogen does not belong to any groups in the periodic table. This is because there are arguments about whether hydrogen belongs to group I. Group I elements are all alkali metals, while hydrogen is not. However, some people says that hydrogen only have one outer shell electron so it should be in group I. Some people even say hydrogen should belong to group VII because it only needs one more electron in order to achieve the duplet of electrons.
Therefore as you may notice, hydrogen in modern periodic tables are put in the center of the periodic table on the top.
Answer:
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
Explanation:
Step 1: Data given
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = Shows how much the boiling point increases
⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate
⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)
⇒m = the molality
Step 2:
0.20 m glucose
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for glucose = 1
⇒ Kb = 0.512 °C/m
⇒m = 0.20 m
ΔT = 1*0.512 * 0.20
<u>ΔT = 0.1024 °C</u>
0.30 m BaCl2
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3
⇒ Kb = 0.512 °C/m
⇒m = 0.30 m
ΔT = 3*0.512 * 0.30
<u>ΔT = 0.4608 °C</u>
0.40 m NaCl
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2
⇒ Kb = 0.512 °C/m
⇒m = 0.40 m
ΔT = 2*0.512 * 0.40
<u>ΔT = 0.4096 °C</u>
0.50 m Na2SO4.
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3
⇒ Kb = 0.512 °C/m
⇒m = 0.50 m
ΔT = 3*0.512 * 0.50
<u>ΔT = 0.768 °C</u>
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
