[ 'm' is greater than or equal to 9 ].There are an infinite number of numbers that satisfy that solution.
Let M equal the gallons of the 90% mixing solution. Let F equal the gallons of the final solution. So:90 + M = F.Also, the number of gallons of pure antifreeze in the final will equal the sum of the gallons of antifreeze in the two mixing parts:Original solution: ( 90 gal )*(0.15) = 13.5 gal [0.15 represents 15%]Mixing solution: M*0.90Final solution: F*0.80So 13.5 + M*0.90 = F*0.80Now you have 2 linear equations and 2 unknowns, you can solve for M & F, using your favorite method: M = 585 and F = 675. Add 585 gallons of the 90% to get 675 gallons of 80% solution.
-log[1 X 10^-4 M OH(-)] = 4 14 - 4 = 10 pH ----------------
The pH is define in the following way: pH = -log [H+] What that means is the pH is the negative of the base 10 logarithm of the concentration of hydrogen ions in the solution. So, if you have a pH = 0, that means that the concentration of H+ is equal to 1 molar, because -log(1) = 0. If you have a 1 M solution of any strong acid, the pH will be equal to zero.
No.
Since HNO3 is a strong acid that completely dissociates in solution, the concentration of H+ ions is equal to the concentration of the acid, which is 0.34 M. The concentration of OH- ions in water is 1.0 x 10^-14 M, therefore, the molarity of OH- ions in the solution is also 1.0 x 10^-14 M.
The concentration of OH- ions in a solution with a pOH of 4.22 is 5.24 x 10^-5 M.
[OH-] = 1x10^-4.22 or more conventionally, [OH-] = 6.03x10^-5 M
The pH of a solution can be calculated using the formula: pH = -log[OH-]. Therefore, for a solution with [OH-] concentration of 10-12 M, the pH would be 12.
The hydroxide ion concentration in a 4.0 x 10^-4 M solution of Ca(OH)2 can be calculated by first finding the molarity of OH- ions from Ca(OH)2, which is twice the molarity of the overall solution. Therefore, the [OH-] is 8.0 x 10^-4 M.
The pH of a 0.0110 M solution of Ba(OH)2 can be calculated by finding the hydroxide ion concentration, which is double the concentration of the Ba(OH)2 solution. Therefore, [OH-] = 2 * 0.0110 M = 0.0220 M. From this, you can calculate the pOH using the formula -log[OH-], and then convert pOH to pH using the relation pH + pOH = 14.
The OH concentration in a 4.0 x 10^4 M solution of Ca(OH)2 can be determined by dividing the concentration of Ca(OH)2 by its stoichiometric coefficient, which is 2. Thus, the OH concentration is 2.0 x 10^4 M.
The pH of a solution can be calculated using the formula pH = 14 - (-log[OH-]). Using the given concentration of 2.3 x 10^-5 M for OH-, the pH would be approximately 9.64.
To find the pH of the solution, first calculate the moles of OH- from each solution (Ca(OH)2 and NaOH). Then, add the moles of OH- together and calculate the total volume of the combined solutions. Finally, use the concentration of OH- ions in the combined solution to calculate the pOH and pH of the solution.
The OH concentration of a 1 M HCl solution is very low because HCl is a strong acid that fully dissociates in water to form H+ ions and Cl- ions. Since it is a strong acid, there is no significant amount of OH- ions present in the solution.
In an oven-cleaning solution with a pH of 12.35, the concentration of hydroxide ions (OH-) would be 1.58 x 10^-1 M. This is calculated by taking the negative logarithm of the hydroxide ion concentration in the solution.
The OH- concentration in a solution can be calculated using the formula OH- = 10^(14 - pH). Therefore, for a solution with a pH of 10.20, the OH- concentration would be 10^(-3.8) M, or approximately 1.58 x 10^(-4) M.