Answer:
B. a strongly basic solution
Explanation:
Kb is base dissociation constant, which indicates how completely a base dissociates into its component ions in water. The greater the Kb value, the greater the alkalinity of the solution and vice versa.
Therefore, a solution with a Kb value much greater than 1, indicates a strongly basic solution, while a solution with a Kb value less than 1, indicates a weakly basic solution.
Answer:
I am sure that the C one is correct
Answer:
a mixture of two these
Explanation:
The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.
n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.
Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.
Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.
Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.
Answer:
21.10g of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2C7H14 + 21O2 —> 14CO2 + 14H2O
From the balanced equation above, 2L of C7H14 produced 14L of H2O.
Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.
Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:
1 mole of a gas occupy 22.4L at stp
Therefore, Xmol of H2O will occupy
26.25L i.e
Xmol of H2O = 26.25/22.4
Xmol of H2O = 1.172 mole
Therefore, 1.172 mole of H2O is produced from the reaction.
Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:
Number of mole H2O = 1.172 mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O =..?
Mass = mole x molar mass
Mass of H2O = 1.172 x 18
Mass of H2O = 21.10g
Therefore, 21.10g of H2O is produced from the reaction.
