<span>Ionic compounds form crystal lattices, not molecules. The term formula unit is used to indicate the simplest whole-number ratio of ions in the compound.</span>
Answer:
Hydrogen bond is formed only by the three highly electronegative elements- fluorine, oxygen and nitrogen.
Answer:
Option c → 8:1
Explanation:
This is the reaction:
C₅H₁₂ + 8O₂ → 10CO₂ + 6H₂O
1 mol of pentane needs 8 moles of oxygen to be combusted and this combustion produces 10 mol of carbon dioxide and 6 moles of water.
To determine the ratio, look the stoichiometry.
For every 8 moles of oxygen, I need 1 mole of pentane gas.
Answer:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
Explanation:
At a substitution reaction by SN1, the alkyl halide must lose its halide, and then an intermediary will be formed: a carbonium, which is an alkyl group with a positive charge in the carbon. The halide lost will be formed the halide ion, which is also an intermediary of the reaction.
The reactivity depends on the stability of the intermediaries (first of the carbonium, and second of the halide ion). As more bonded with carbons is the carbonium, more stable it is. The order of stability of the halides ions is from their electronegativity: as lower is it, as stable is the ion. The order is then: I⁻ > Br⁻ > Cl⁻ > F⁻.
2-bromo-2-methylpentane, 2-chloro-2-metylpentane, and 2-iodo-2-methylpentane, will form a 3-degree intermediary, so they will be more reactive than 3-chloropentane, which form a 2-degree intermediary. So, for the order of the stability of the halide ions, the order of reactivity is:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
Answer:
1.72 moles of H₂
Explanation:
The balanced equation
2Na+ 2H₂O → <u>2</u>NaOH + <u>1C</u>
tells us we'll get <u>1</u> mole of H₂ for every <u>2</u> moles of NaOH. We can express this as a molar ratio: (1 mole H₂)/(2 moles NaOH).
The reaction produces 138 g of NaOH. Divide this mass by the molar mass of NaOH (39.99 g/mole)to obtain moles NaOH:
138g NaOH/(40 g/mole) = 3.45 moles of NaOH.
To find the moles H₂ produced, we can multiply the moles NaOH by the molar ratio from above:
(3.45 moles NaOH)*((1 mole H₂)/(2 moles NaOH) = (3.45/2) moles H₂. [Note how the "moles NaOH" unit cancels, leaving just "moles H₂"]
We should produce 1.72 moles of H₂.
