Answer: acetone molecule ( CH₃-CO-CH₃)
Explanation:
1) Acetone is CH₃-CO-CH₃
2) That is a molecule (build up of covalent bonds).
3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.
This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.
That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).
Answer:
For these types of questions the equation that we must take into account is that:
T = PxV (where T is the temperature, P is the pressure and V is the volume) this equation is described as we consider that this is the value N and R is 1, therefore it is not necessary to explain them now.
Explanation:
The quoted equation refers to Boyle's Law, in this law we can explain that the volume increases if the pressure decreases and if the temperature also increases, if the pressure increases and the volume decreases this means that the gas is compressing assuming that the temperature is constant
The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: <span>[He] 2s</span>²<span> 2p</span>⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. <em>Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.</em>
The balanced equation for the above reaction is as follows;
<span>10Na(s) + 2KNO</span>₃(s)--> K₂O(s) + 5Na₂O(s) + N₂<span>(g)
Stoichiometry of Na to KNO</span>₃ is 10 : 2
Number of Na moles reacted - 5.1 g/ 23 g/mol = 0.22 mol
Na is the limiting reactant, therefore Na is fully used up, Since KNO₃ is present in excess , at the end of the reaction a certain amount of KNO₃ will be remaining.
10 mol of Na reacts with 2 mol of KNO₃
Therefore 0.22 mol of Na reacts with - 2 /10 x 0.22 = 0.044 mol of KNO₃
Mass of KNO₃ reacted = 0.044 mol x 101.1 g/mol = 4.45 g
Mass of KNO₃ present initially - 305 g
Therefore remaining mass of KNO₃ - 305 - 4.45 = 300.55 g
You can use this formula to solve for density--> Density= PM/ RT, where P is pressure, M is molar mass, R is the gas constant and T is temperature.
P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k
density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
