Reaction: 2K₍s₎ + 2H₂O₍l₎ → 2KOH₍aq₎ + H₂₍g₎.
K - potassium.
H₂O - water.
KOH - potassium-hydroxide.
H₂ - hydrogen.
s - solid phase.
l - liquid.
aq - disolves in water.
g - gas.
Reaction is exothermal (release of energy) and potassium burns a purple flame. H<span>ydrogen released during the reaction reacts with </span>oxygen<span> and ignites.</span><span>
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First you calculate the concentration of [OH⁻] in <span>solution :
POH = - log [ OH</span>⁻]
POH = - log [ 0.027 ]
POH = 1.56
PH + POH = 14
PH + 1.56 = 14
PH = 14 - 1.56
PH = 12.44
hope this helps!
Answer: 8500 J lost or -8500 J
Explanation:
q=cmt
t=375-26
t=349
q=0.38(64)(349)
q=8487.68 J
Answer must have 2 sig figs, which means it rounds to 8500 J.
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