Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
pH = 8.314
Explanation:
equil: S S 3S
∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24
⇒ 6.0 E-24 = ( S )*( 3S )³
⇒ 6.0 E-24 = 27S∧4
⇒ 2.22 E-25 = S∧4
⇒ ( 2.22 E-25 )∧(1/4) = S
⇒ S = 6.866 E-7 M
⇒ [ OH- ] = 3*S =2.06 E-6 M
⇒ pOH = - Log [ OH- ]
⇒ pOH = - Log ( 2.06 E-6 )
⇒ pOH = 5.686
∴ pH = 14 - pOH
⇒ pH = 8.314
<span>UV gel enhancements rely on ingredients from the monomer liquid and polymer powder chemical family. The chemicals from the polymer powder family </span><span>can absorb and retain extremely large amounts of a liquid relative to their own mass. Water-absorbing polymers, </span>can absorb aqueous solutions through hydrogen bonding with water molecules.
