I believe it is d. scientific law
Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Answer:
iodine
Explanation:
In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.
<u>Answer:</u>
<em>0.264 g of 
can be formed from 288 mg of 
</em>
<u>Explanation:</u>
The balanced chemical equation is
The conversions are
Mass in mg is converted to mass in g
Mass in g is converted to moles by dividing with molar mass
Moles is converted to moles by using the mole ratio of 
is 9 : 6
Moles is converted to mass by multiplying with molar mass
mass in mg > mass in g >moles > moles > mass
=0.264g (Answer)
Explanation:
I have a dog in my dog is a girl
