An atom consists of a nucleus of protons and neutrons, surrounded by electrons. Each of the elements in the periodic table is classified according to its atomic number, which is the number of protons in that element's nucleus. Protons have a charge of +1, electrons have a charge of -1, and neutrons have no charge. Neutral atoms have the same number of electrons and protons, but they can have a varying number of neutrons. Within a given element, atoms with different numbers of neutrons are isotopes of that element. Isotopes typically exhibit similar chemical behavior to each other.
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.
For the Lewis diagram of the cyanide ion, a
figure is shown.<span>
There are 3 pairs of bonding electrons. There is a
one lone pair each for the carbon and the nitrogen atoms.</span>
<span>I hope I was able to answer your question. Thank
you!</span>
First lets do Rb..we have 1 one the left and 3 on the right. so 3RbOH will balance that
PO4 is equal on both side
since we put 3 for RbOH now we have 3 O on the left
so we put 3 for H2O
now we have 6 H on each side so everything is balanced