The balanced chemical reaction is given as follows:
<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)
The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:
</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
C=10⁻⁶ mol/L
pH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg10⁻⁶=14-6=8
B. pH = 8
<span>The mixture that is most likely to form a suspension is flour and liquid water mixed together, as in a mixture like gravy. A suspension mixture is a mixture that has large solid particles, particles that are large enough for sedimentation.</span>
Answer:
Copper ions are reduced into copper atoms.
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
Explanation:
During electrolysis, the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.
At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)
Thus the half equation is as follows:
Cu²⁺₍aq₎ + 2e⁻ → Cu₍s₎
