44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3
46. H Br
δ+ δ−
48. The metallic potassium atoms lose one electron and form +1 cations,
and the nonmetallic fluorine atoms gain one electron and form –1 anions.
K → K+
+ e–
19p/19e–
19p/18e–
F + e–
→ F–
9p/9e–
9p/10e–
The ionic bonds are the attractions between K+
cations and F–
anions.
50. See Figure 3.6.
52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal
54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic
56. (a) 7 (b) 4
58. Each of the following answers is based on the assumption that nonmetallic
atoms tend to form covalent bonds in order to get an octet (8) of
electrons around each atom, like the very stable noble gases (other than
helium). Covalent bonds (represented by lines in Lewis structures) and lone
pairs each contribute two electrons to the octet.
(a) oxygen, O
If oxygen atoms form two covalent bonds, they will have an octet of electrons
around them. Water is an example:
H O H
(b) fluorine, F
If fluorine atoms form one covalent bond, they will have an octet of electrons
around them. Hydrogen fluoride, HF, is an example:
H F
(c) carbon, C
If carbon atoms form four covalent bonds, they will have an octet of electrons
around them. Methane, CH4, is an example:
H H
H
H
C
(d) phosphorus, P
If phosphorus atoms form three covalent bonds, they will have an octet
I think it is feeling warm air while standing in front of the blower from a heating system.
Hope this helps! :)
Answer:they are used to generate sound waves in the ultrasonic range
Explanation:transducers are used to generate sound waves in the ultrasonic range ,by turning electrical energy into sound
I'm pretty sure its metals that make good conductors.
t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half life of the element and λ is decay constant.
32 = 0.693 / λ
λ = 0.693 / 32 (1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, λ is decay constant and t is the time taken.
t = 1.9 hours = 1.9 x 60 min
From (1) and (2),
Nt = Nο e⁻Λ(0.693/32)*1.9*60
Nt = 0.085Nο
Percentage = (Nt/Nο) x 100%
= (0.085Nο/Nο) x 100%
= 8.5%
Hence, Percentage of remaining atoms with the original sample is 8.5%
