Adding solvent or removing solute from a solution is called diluting. And a solution is said to be concentrated if it has more solute. The opposite of diluting is called concentrating. The measure of the amount of solute in a solution is expressed in concentration.
Do length x width x height which is 10 cm x 8.2 cm and 3.5 cm. Pay close attention to sig figs as well (or if your teacher doesn't mind all that much then don't fret about it, but mine's really picky!)
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
The answer is B....Brainliest?....
Hello!
The dissociation reaction for Benzoic Acid is the following:
C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺
The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29](https://tex.z-dn.net/?f=Ka%3D%20%5Cfrac%7B%5BC_6H_5COO%5E%7B-%7D%5D%2A%5BH_3O%5E%7B%2B%7D%5D%20%7D%7B%5BC_6H_5COOH%5D%7D%3D%5Cfrac%7BX%2AX%20%7D%7B0%2C40%20-X%7D%20%20%28assume%3A%200%2C40-X%5Capprox0%2C40%29%5C%5C%20%20%5C%5C%20X%3D%20%5Csqrt%7B0%2C40%2A6%2C50%2A10%5E%7B-5%7D%20%7D%3D0.00510M%20%5C%5C%20%20%5C%5C%20pH%3D-log%28%5BH_3O%5E%7B%2B%7D%5D%3D2%2C29%20%20%20)
So, the pH of this Benzoic Acid solution is
2,29Have a nice day!