An acid a. has a high pH in solution.
b. turns blue litmus paper to red.
c. releases hydroxyl ions in solution.
d. has more hydroxyl than hydrogen (or hydronium) ions.
<span>An acid </span>turns blue litmus paper to red. The answer is letter B.
Explanation:
The given data is as follows.
(NaCl) =
(H-O=C-ONO) =
(HCl) =
Conductivity of monobasic acid is
Concentration = 0.01
Therefore, molar conductivity () of monobasic acid is calculated as follows.
=
=
=
Also, =
=
=
Relation between degree of dissociation and molar conductivity is as follows.
=
= 0.1254
Whereas relation between acid dissociation constant and degree of dissociation is as follows.
K =
Putting the values into the above formula we get the following.
K =
=
=
=
Hence, the acid dissociation constant is .
Also, relation between and is as follows.
=
= 3.7454
Therefore, value of is 3.7454.
Examination by a test; experiment, as in chemistry, metallurgy, etc.
Answer:
Option C = object B by 1 gram per cubic cm.
Explanation:
Given data:
Mass of object A = 12 g
Volume of object A = 8 cm³
Mass of object B = 20 g
Volume of object B = 8 cm³
Densities = ?
Solution:
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
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Density of object A:
d = m/v
d = 12 g/ 8 cm³
d = 1.5 g/cm³
Density of object B:
d = m/v
d = 20 g/ 8 cm³
d = 2.5 g/cm³
object b has high density.
Ions and charges= Al+3 SO3-2
The charges must add up to zero, in order for two +3 aluminum ions to cancel out three -2 sulfite ions.
Al+3 Al+3 SO3-2 SO3-2 SO3-2
<span>So simply put : </span>Al2(SO3)<span>3 </span>