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:
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Explanation:
The answer is C as enzymes are biological catalysts that act on a substrate such as starch if the enzyme were amylase
hope that helps
For Row 1 according to the respective Column is
- Zn^{2+} , 32 ,65 ,32 ,30 ,34 ,2+
For Row 2 according to the respective Column is
- Mn, 25, 55, 25, 25,30, 3-
For Row 3 according to the respective Column is
For Row 4 according to the respective Column is
- O^{2-} , 8 ,16 ,8 ,6 ,8 , 2-
For Row 5 according to the respective Column is
- Na, 11, 23, 11, 10, 12, 1-
From the question we are told
Ions, isotope, average atomic mass fill in the blank
Generally
Where
- Z signifies the Proton Number of the atom
- A signifies Mass Number of the atom
Therefore
For Row 1 according to the respective Column is
- Zn^{2+} , 32 ,65 ,32 ,30 ,34 ,2+
For Row 2 according to the respective Column is
- Mn, 25, 55, 25, 25,30, 3-
For Row 3 according to the respective Column is
For Row 4 according to the respective Column is
- O^{2-} , 8 ,16 ,8 ,6 ,8 , 2-
For Row 5 according to the respective Column is
- Na, 11, 23, 11, 10, 12, 1-
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