Chemical equations must always balance due to the principles outlined in The Law Of Conservation of Matter. This scientific law states that matter cannot be created out of nothing nor can it be destroyed.
should be ocean ridges im not sure hope this helps tho
Answer:the pH is 12
Explanation:
First We need to understand the structure of trimethylamine
Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline
C3H9N+ H2O --> C3H9NH + OH-
![k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC3H9NH%5D%2A%5BOH-%5D%7D%7B%5BC3H9N%5D%7D)
Then:
The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine
[OH-]=
[OH-]=0.01
pH=14-(-log[OH-])
pH=12
Answer:
0.17%
Explanation:
With the equation:
2Cr2O7 2- + C2H5OH + H2O --> 4Cr3+ + 2CO2 + 11H2O
We can assume that every mole of ethanol needs 2 moles of Dichromate to react.
So if in 1L we have 0.05961 moles of dichromate we can discover how many moles we have in 35.46mL
1000 mL - 0.05962 moles
35.46 mL - x
x = 
x = 2,11* 10^-3 moles
As we said earlier, 1 mole of ethanol needs 2 mole of dichromate, so in the solution we have 1,055*10^-3 moles of ethanol. We can discover the mass of ethanol present in the solution.
1 mole - 46g
1.055*10^-3 - y
y = 46 * 1.055*10^-3
y = 0.048 g
To discover the percent of alchol we can use a simple relation
28 g - 100%
0.048 - z
z = 
z = 0.17%
Answer:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Explanation:
The balanced reaction between Na2CO3 and HCl is given as;
Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)
The next step is o express the species as ions.
The complete ionic equation for the above reaction would be;
2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)
The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.
The spectator ions are; Na⁺ and Cl⁻
The net ionic equation is given as;
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)