Compete Question:
What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?
Passage: "16 mmol of CDP in 1 L of buffer"
Answer:
6.4 × 10-2 g
Explanation:

we are given from the question that 16 mmol of CDP is in 1 L of buffer
this mean that we have
moles of CDP in 1 liter of buffer.
so the mass of CDP in one liter of buffer will be calculate as,
mass of CDP =
× 403g mol−1
=
= 6.4 g/L
But because the question
asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:
6.4 g/L × 10 mL
6.4 g/L × 0.01 L =
Answer:
8.32 s⁻¹
Explanation:
Given that:
The concentration of myosin = 25 pmol/L
R_max = 208 pmol/L/s
The objective is to determine the turnover number of the enzyme molecule myosin, which has a single active site.
In a single active site of enzyme is known to be a region where there is binding of between substrate molecules, thereafter undergoing chemical reaction.
The turnover number of the enzyme is said to be the number of these substrate molecule which binds together are being converted into products.
The turnover number of the enzyme molecule of myosin can be calculated by the expression: 
⇒ 
= 8.32 s⁻¹
Answer:
The nucleus consists of 9 protons and 10 neutrons. Nine electrons occupy available electron shells.
Molarity is moles over liter. When checking for molarity, before diving, make sure the units are correct. In this case, both units are moles and liters, so we can assume the molarity is 3/12, or 0.25 M