Answer:
D) 2-methylpent-2-ene
Explanation:
This is an elimination reaction of Halogenoalkane. 2-bromo-2-methylpentane when is heated with NaOH or NaOC2O5( sodium ethoxide) in ethanol will form alkene rather than alcohol.
2-methylpent-1-ene is minor product since double bond form with secondary Carbon rather than primary Carbon.
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
<span>These atoms are known as valence atoms.</span>
Answer:
B. Changes in energy in systems.
Explanation:
Thermodynamics is a branch of physics which deals with heat and temperature in relation to energy work radiation and characteristics of matter.
This field studies the causes of energy changes and the effects of energy changes to chemical and physical systems.
Answer is: the hydrogen ion concentration is 10⁻¹⁰ mol/L, solution is basic.
[OH⁻] =0.00001 mol/L = 10⁻⁴ M.
[OH⁻]·[H⁺<span>] = Kw.
</span>0.00001 mol/L ·[H⁺] = 10⁻¹⁴ mol²/L².
[H⁺] = 10⁻¹⁴ mol²/L²÷ 10⁻⁴ mol/L.
[H⁺] = 10⁻¹⁰ mol/L.
pH = -log[H⁺].
pH = -log(10⁻¹⁰ mol/L).
pH = 10.
If pH is less than seven, solution is acidic; greater than seven, solution is basic; equal seven, solution is neutral.
