D. 6 protons and 7 neutrons
As proton remains constant
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 %
The molecular formula for hyponitrous acid is H2N2O2. and for nitroxyl is HNO.
The chemical compound HNO is also known as nitroxyl (common name) or Azanon (IUPAC name). In the gas phase, it is widely recognized. In the solution phase, the short-lived intermediate nitroxyl can develop. Nitric oxide (NO) is reduced to form the conjugate base, NO, which is isoelectronic with dioxygen.
By oxidizing hydroxylamine with CuO , HgO, and Ag 2 and by oxidizing hydroxylamine with N2O3 in methyl-alcoholic solution, we can create hyponitrous acid.
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