n = m / M
Where, n is moles of the compound (mol), m is the mass of the compound (g) and M is the molar mass of the compound (g/mol)
Here, the given ethanol mass = 50.0 kg = 50.0 x 10³ g
Molar mass of the ethanol = (12 x 2 + 1x 6 + 1 x 16) g/mol
= 46 g/mol
Hence, moles in 50.0kg of ethanol = 50.0 x 10³ g / 46 g/mol
= 1086.96 mol
Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.<span>
Chemical reaction: HCOOH(aq) + H</span>₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).<span>
c(HCOOH) = 0,1 M.
[</span>H₃O⁺] = [HCOO⁻] = x.<span>
[HCOOH] = 0,1 M - x.
</span>Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).<span>
Solve quadratic equation: x = </span>[H₃O⁺] = 0,0042 M.
<span>The bonding found in calcium chloride is i</span>onic bonds.
I hope this helps!
Answer:
The pH is equal to 4.41
Explanation:
Since HClO is a weak acid, its dissociation in aqueous medium is:
HClO ⇄ ClO- + H+
start: 0.05 0 0
change -x +x +x
balance 0.05-x x x
As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.
the acidity constant when equilibrium is reached is equal to:
![Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BClO-%5D%2A%5BH%2B%5D%7D%7B%5BHClO%5D%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.05-x%7D%3D3x10%5E%7B-8%7D)
The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:
clearing the x and calculating its value we have:
![x=3.87x10^{-5}=[H+]=[ClO-]](https://tex.z-dn.net/?f=x%3D3.87x10%5E%7B-5%7D%3D%5BH%2B%5D%3D%5BClO-%5D)
the pH can be calculated by:
![pH=-log[H+]=-log[3.87x10^{-5}]=4.41](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D%3D-log%5B3.87x10%5E%7B-5%7D%5D%3D4.41)
C. a change of state. It can be a physical or a chemical state of change.
