Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.
1 answer:
Answer: 45.983 g CBr₄
Explanation:
To convert from moles to grams, you know that we will need molar mass and Avogadro's number.
Avogadro's number: 6.022×10²³ molecules/mol
Molar mass: 331.627 grams/mol
Now that we have what we need, you can use these to solve for grams.
Our final answer is 45.983 g CBr₄.
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Answer:
760 uM
Explanation:
<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>
The problem here is that the amount of magnesium ion remains the same irrespective of the volume.
Amount of magnesium in the growth medium = <em>molarity x volume</em>
= 97 x x 47 x
= 4.559 x
Then, the volume reduced to 6.0 mL, the new molarity becomes;
<em>molarity = mole/volume </em>
= 4.559 x /6 x
= 7.598333 x
M = 759.83333 uM
To the correct number of significant digits = 760 uM
Here’s the answer to your question :)
