Answer:
not 100% but i think its 1.57x10^20
Explanation:
5.25x10^-4g / 2.016g
2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules
Answer:
![\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B1.447%20%5Ctimes%2010%5E%7B23%7D%5Ctext%7B%20molecules%20Cu%28OH%29%7D_%7B2%20%7D%7D)
Explanation:
1. Calculate the moles of copper(II) hydroxide
![\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Cu%28OH%29%7D_%7B2%7D%20%3D%20%5Ctext%7B23.45%20g%20Cu%28OH%29%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20Cu%28OH%29%7D_%7B2%7D%7D%7B%5Ctext%7B97.562%20g%20Cu%28OH%29%7D_%7B2%7D%7D%20%3D%20%5C%5C%5C%5C%5Ctext%7B0.240%2036%20mol%20Cu%28OH%29%7D_%7B2%7D)
2. Calculate the molecules of copper(II) hydroxide
<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D%7B%5B%5Ctext%7Bacid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
= moles of conjugate base = 3.25 moles
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:
![pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61](https://tex.z-dn.net/?f=pH%3D4.70%2B%5Clog%28%5Cfrac%7B3.25%7D%7B4.00%7D%29%5C%5C%5C%5CpH%3D4.61)
Hence, the pH of the buffer is 4.61
Crystals of hydrated magnesium sulfate used as a purgative or for other medicinal use
It on googIe
Hope that helps