Answer:
They are both listed under group 11 on the periodic table and both are highly conductive of electricity
Explanation:
HOPE THIS HELPS ^^
Answer:
91%
Explanation:
Since you are just trying to find the yield, take the moles of the substance from the products and divide it by the mol value of the reactants. Multiply by 100 to find percentage
0.45 ÷ 0.41 = 0.91111...
91%
Answer:
A. 8.4
Explanation:
[OH⁻] = 2.6 × 10⁻⁶ Take the negative log of each side
-log[OH⁻] = pOH = 5.59 Apply the pH/pOH relation
pH + pOH = 14.00 Insert the value of pOH
pH + 5.59 = 14.00 Subtract 5.59 from each side
pH = 14.00 -5.59 = 8.41