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 = 
Is this a multiple choice?
Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.
Explanation :
According to dilution law:
where,
= molarity of aqueous sodium carbonate
= molarity of aqueous sodium carbonate stock solution
= volume of aqueous sodium carbonate
= volume of aqueous sodium carbonate stock solution
Given:
= 1.00 M
= 1.58 M
= 575 mL
= ?
Now put all the given values in the above formula, we get:
Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.
Answer:
Different atmospheric pressure. When there is a different atmospheric pressure, air moves from the higher pressure to the lower pressure area which results in what you call <u>WIND</u> but can result in various speeds and pressure.
Hope this helped and if it did, please give my answer a brainliest.
18.is D
19.is A
hope it's correct
