Since the acid and base react in a one to one ratio, you can use the equation M1V1 = M2V2 to solve for M2, or the molarity of NaOH.
M1 = 0.9 M HCl
V1 = 200.0 mL HCl
M2 = ?
V2 = 23.9 mL NaOH
(0.9 M)(200.0 mL) = M2(23.9 mL) —> M2 = 7.53 M NaOH
Answer: i dont understand
Explanation:
Answer:- The pressure of ethanol would be 109 mmHg.
Solution:- This problem is based on Clausius clapeyron equation--
Given, = 63.5 + 273 = 336.5 K
= 34.9 + 273 = 307.9 K
= 400 mmHg
= ?
= 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/mol.K
Let's plug in the values in the equation and do the calculations.
= 1.30
On taking anti ln to both sides...
=
= 3.67
= 400/3.67
= 109 mmHg
To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.
Let us determine the reaction for the titration below:
2NaOH +2H2SO4 = Na2SO4 +2H2O
So,
0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution
= 2.62 M H2SO4
The answer is the fourth option:
<span>2.62 M</span>
Answer : Option B) When the equilibrium constant is large.
Explanation : The expression for an equilibrium constant is expressed as the product of the concentrations of the reaction products in the numerator and the product of the concentrations of the reactants in the denominator.
Each of the concentrations are raised to the power of their respective stoichiometric coefficients in the reaction.
When the equilibrium is large the products are favored over the reactants, as concentration of products are directly proportional to the equilibrium constant.
The expression of equilibrium constant is given by
K =
where K - equilibrium constant;
a - co-efficient of product; and
b - co-efficient of reactant.