Answer:
The answer to your question is 16 g
Explanation:
Data
Percent by mass = 8%
Mass of the solution = 200 g
Mass of solute = ?
Formula
Percent by mass = mass of solute / mass of solution x 100
- Solve for mass of solute
Mass of solute = Percent by mass x mass of solution / 100
- Substitution
Mass of solute = 8 x 200 / 100
- Simplification
Mass of solute = 1600 / 100
- Result
Mass of solute = 16 g
Answer:
It's Effective Collision.
Explanation:
Hope my answer has helped you!
<span>136.14 g/mol </span><span><span>Calcium sulfate, Molar mass</span></span>
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
Answer:
Sodium laurate, also known as sodium dodecanoate, is a soap. It is the salt of lauric acid. It is an amphiphilic organic molecule which is composed of a hydrophilic head (polar ) and a hydrophobic tail (non-polar fatty acid).
In a aqueous solution, it leads to the formation of a micelle. The hydrophilic head of the molecule interacts with the surrounding polar solvent molecules. Thereby, making the micelle soluble in the solution. Whereas, the hydrophobic tails present in the core of micelle, interacts with the non-polar oil particles.
