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>
</span>
No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
