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
Answer:
Explanation:
Hello there!
In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:
Thus, by solving for the volume of the base as NaOH, we obtain:
Best regards!
To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:
r1/r2 = √(m2/m1)
where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol)
r1 = 1.6r2
1.6 = √(44.01 / m1)
m1 = 17.19 g/mol
Answer:
2 moles
Explanation:
In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams
