Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻.

pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

pH + pOH = 14

pH + 3.30 = 14

pH = 14 - 3.30

pH = 10.7

hope this helps!

5 0

3 years ago

How do you solve this ??

kotegsom [21]

Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

$E_{eV} = E_{J} X (6.241 X 10^{18})$

Here, we have the value of particle in terms of Joules which is 3.2 X $10^{-19}$

So, on substituting we get,
$E_{eV}$ = 3.2 X $10^{-19}$ X $(6.241 X 10^{18} )$

$E_{eV}$ = 1.99 eV so, it can be rounded off to 2.00 eV.

3 0

3 years ago

What is the element present for copper

amm1812

Copper is an brown-orange color which it's atomic number is 29. With high thermal and electricity conductivity with it's smooth surface.

8 0

2 years ago