Answer:
35.9 ml
Explanation:
Start with the balanced equation:
3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-
∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4
We know that concentration = moles/volume i.e:
c= n/v
∴n=c×v
∴nCuCl2=0.107×91.01000=9.737×10−3
I divided by 1000 to convert ml to L
∴nNa3PO4=9.737×10−3×23=6.491×10−3
v=nc=6.491×10−30.181=35.86×10−3L
∴v=35.86ml
The more protons you add, the more positively charged the atom becomes
the charge of an atom determines what kind of an atom it is <span />
Answer:
Q was < K. Partial pressure of hydrogen decreased, iodine increased
Explanation:
<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>
Based on the equilibrium:
H2(g) + I2(g) ⇄ 2HI(g)
K of equilibrium is:
K = [HI]² / [H2] [I2]
<em>Where [] are concentrations at equilibrium</em>
And Q is:
Q = [HI]² / [H2] [I2]
<em>Where [] are actual concentrations of the reactants.</em>
<em />
When the reaction is in equilibrium, K=Q.
But as [I2] is increased, Q decreases and Q was < K
The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased
