The pH value of the solution is mathematically given as
pH=2.35
<h3>What
pH value of the
solution?</h3>
Question Parameters:
pH during the titration of 20.00 mL of 0.1000 M dimethylamine,
with 0.1000 M HCl(aq) after 21.23 mL of the acid
Generally, the equation for the Chemical Reaction is mathematically given as
(CH3)2NH(aq), +Hcl ---> <---- (CH3)2NH2Cl(aq)
Therefore
HCL=0.00444M
WHere
HClaq--->H+(aq)+Cl-(aq)
Hence
H+=0.00444M
pH= -log{H+}
pH=log(0.00444)
pH=2.35
For more information on Chemical Reaction
brainly.com/question/11231920
Answer:
A , B, C
Explanation: D is a Diamagnetic
Hey there!
Balance the equation:
SiCl₄ + H₂O → H₄SiO₄ + HCl
Balance H.
2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.
SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl
Balance O.
3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl
This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Balance Cl.
4 on the left, 4 on the right. Already balanced.
Balance Si.
1 on the left, 1 on the right. Already balanced.
Our final balanced equation:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Hope this helps!
Answer:
lower mantle
Explanation:
because the core and the crust are solid.
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
