Answer:
Any kind, as long as there is an action.
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry) 
= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
The amount of energy released when 0.06 kg of mercury condenses at the same temperature can be calculated using its latent heat of fusion which is the opposite of melting. Latent heat of fusion and melting can be used because they have the same magnitude, but opposite signs. Latent heat is the amount of energy required to change the state or phase of a substance. For latent heat, there is no temperature change. The equation is:
E = m(ΔH)
where:
m = mass of substance
ΔH = latent heat of fusion or melting
According to data, the ΔH of mercury is approximately 11.6 kJ/kg.
E = 0.06kg (11.6 kJ/kg) = 0.696 kJ or 696 J
The answer is D. 697.08 J. Note that small differences could be due to rounding off or different data sources.
Answer:
The atomic radius corresponds to Sigma
Explanation:
It is the Van Der Waals radius ;)
K because parent atoms are always larger than their cations(positively charged atoms)
