Answer:
A
Explanation:
Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.
As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain
1) Dawn dish soap has a density of 1.06 g/mL. If the mass of a sample of the liquid is 1.00 g what is the volume?
Answer:
v = 0.94 mL
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Given data:
Density of soap = 1.06 g/mL.
Mass = 1 g
Volume = ?
Solution:
d = m/v
v = m/d
v = 1 g/1.06 g/mL
v = 0.94 mL
2) Maple syrup has a density of 1.37 g.mL. What is the mass of 1.0 L of the maple syrup?
Answer:
m = 1370 g
Given data:
Density of soap = 1.37 g/mL.
Mass = ?
Volume = 1.0 L ( 1000 mL)
Formula:
D=m/v
D= density
m=mass
V=volume
Solution:
d = m/v
m = d × v
m = 1.37 g/mL × 1000 mL
m = 1370 g
3) The density of gasoline is 0.754 g/mL. A drop of gasoline has a mass of 22 g what is the volume?
Answer:
v = 29.2 mL
Given data:
Density of soap = 0.754 g/mL.
Mass = 22 g
Volume = ?
Formula:
D=m/v
D= density
m=mass
V=volume
Solution:
d = m/v
v = m/d
v = 22 g/0.754 g/mL
v = 29.2 mL
Answer:Ionic compounds form when positive and negative ions share electrons and form an ionic bond. ... The positive ion, called a cation, is listed first in an ionic compound formula, followed by the negative ion, called an anion. A balanced formula has a neutral electrical charge or net charge of zero.
Explanation:Simple ions:
Perchlorate ClO4- IO3-
Chlorate ClO3- BrO3-
Chlorite ClO2-
Hypochlorite OCl- OBr-
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
